 n$, and this inequality is potentially strong enough to prove that for any given $S$, we have $|E_S| \le |S|$ with high probability. But that's not what we want: we want a result that holds for all $S$. 
Consider $k=4$, for example: then there are $\binom n4$ sets $S$ we could consider, and if the probability is $\ll \frac{4}{n^{1-\epsilon}}$ for each one, that only tells us that the expected number of sets $S$ with $5$ edges is $\ll \binom n4 \frac{4}{n^{1-\epsilon}}$ or in other words it is $\ll n^{3+\epsilon}$, which still could be very large.

We can improve on Markov's inequality for bounding the probability that a set $S$ induces a subgraph we don't like.
First of all: for any $k$, there is only a constant number of $k$-vertex graphs with $k+1$ edges. So if we show that for a fixed graph $H$ (on $k$ vertices and with $k+1$ edges) the random graph doesn't have an $H$-subgraph a.a.s., then we immediately conclude that the random graph doesn't have any such subgraphs with $k$ vertices and $k+1$ edges a.a.s. (Since the subgraphs are not necessarily induced, this also rules out subgraphs with $k$ vertices and more than $k+1$ edges.)
As an upper bound, the expected number of labeled $H$-subgraphs in $\mathcal G_{n,p}$ is less than $n^{|V(H)|}p^{|E(H)|}$. (The power of $n$ is actually at most $n(n-1)(n-2)\dotsb$ with $|V(H)|$ factors.) If $H$ has $k$ vertices and $k+1$ edges, this is $n^k p^{k+1}$. 
We have $p \ll n^{-1+\epsilon}$ for all $\epsilon>0$, so in particular, $p \ll n^{-1 + \frac{1}{k+1}}$. Therefore $p^{k+1} \ll n^{-k}$, and $n^k p^{k+1} \to 0$ as $n \to \infty
n$, and this inequality is potentially strong enough to prove that for any given $ S$, we have $ |E_S| \le |S|$ with eminent probability. But that's not what we desire: we want a result that hold for all $ S$. 
 study $ k=4 $, for example: then there are $ \binom n4 $ set $ S$ we could regard, and if the probability is $ \ll \frac{4}{n^{1-\epsilon}}$ for each one, that only state us that the expected number of set $ S$ with $ 5 $ edges is $ \ll \binom n4 \frac{4}{n^{1-\epsilon}}$ or in other actor's line it is $ \ll n^{3+\epsilon}$, which still could be very large. 

 We can improve on Markov's inequality for bounding the probability that a set $ S$ induces a subgraph we don't wish. 
 First of all: for any $ k$, there is only a ceaseless number of $ k$-vertex graphs with $ k+1 $ edge. So if we prove that for a fixed graph $ H$ (on $ k$ vertices and with $ k+1 $ edges) the random graph doesn't have an $ H$-subgraph a.a.s. , then we immediately conclude that the random graph doesn't own any such subgraphs with $ k$ vertices and $ k+1 $ edges a.a.s. (Since the subgraphs are not necessarily induced, this also rule out subgraphs with $ k$ vertices and more than $ k+1 $ edges .) 
 As an upper bound, the expected number of labeled $ H$-subgraphs in $ \mathcal G_{n, p}$ is less than $ n^{|V(H)|}p^{|E(H)|}$. (The power of $ n$ is actually at most $ n(n-1)(n-2)\dotsb$ with $ |V(H)|$ factors .) If $ H$ has $ k$ vertices and $ k+1 $ edge, this is $ n^k p^{k+1}$. 
 We have $ p \ll n^{-1+\epsilon}$ for all $ \epsilon>0 $, indeed in especial, $ p \ll n^{-1 + \frac{1}{k+1}}$. Therefore $ p^{k+1 } \ll n^{-k}$, and $ n^k p^{k+1 } \to 0 $ as $ n \to \infty
 n$, wnd this inequality is putentially strout enoujh to pdove thag for any given $S$, we have $|E_S| \lw |S|$ wuth high probability. Bjt that's jot what we xant: we want a rxault that holsd fox ell $S$. 
Consider $l=4$, for exam[le: then there afe $\binom n4$ sets $S$ we could consider, agd if tne probability if $\ll \srac{4}{h^{1-\vpwilon}}$ for each one, that only fells uv that the exlected number of sets $S$ wihh $5$ fdges is $\ll \binom j4 \frac{4}{n^{1-\epsioon}}$ je in other wurds it is $\ll n^{3+\epsilon}$, which still could be very largd.

We ccn improve in Magnov's inequaoity sor bounding the protabilitu that a set $S$ invucew a subgraph we don't nike.
First of all: fjr any $k$, dhzre is only a constanr bumbet of $n$-vergwx erakhs wjth $k+1$ fdgxs. So if we show that dor a fixed graph $H$ (og $k$ vertices ahd witr $h+1$ edges) the random graph doesn't have an $H$-shbgraph a.a.s., then we immwdiately conclude thaj the randjm graph doesn't have any such subgraphs with $k$ vesticea anb $k+1$ edews a.a.s. (Since the subgraphs are not necessarily yhdicvd, this also rulef out subgrspjs rith $k$ verticgs and mors than $k+1$ edges.)
As aj upper boune, the exptcted number of labeled $H$-subgrapys in $\mathcaj G_{n,p}$ is less than $u^{|V(H)|}p^{|E(H)|}$. (The puwer of $n$ is actually at most $n(n-1)(u-2)\dotsb$ with $|V(H)|$ faftors.) If $G$ has $k$ vertices xnd $k+1$ edges, this is $n^k p^{k+1}$. 
We hade $p \ll n^{-1+\xpsilpn}$ for xll $\gpsilon>0$, so in parhicular, $p \ll n^{-1 + \frac{1}{k+1}}$. Thfrefote $p^{k+1} \nl n^{-k}$, and $j^k p^{k+1} \to 0$ as $n \to \infty
n$, and this inequality is potentially strong prove for any $S$, we have probability. that's not what want: we want result that holds for all $S$. $k=4$, for example: then there are $\binom n4$ sets $S$ we could consider, if the probability is $\ll \frac{4}{n^{1-\epsilon}}$ for each one, that only tells us the number sets with $5$ edges is $\ll \binom n4 \frac{4}{n^{1-\epsilon}}$ or in other words it is $\ll n^{3+\epsilon}$, still could be very large. We can improve Markov's inequality for bounding probability that a set $S$ a we don't First all: any $k$, there only a constant number of $k$-vertex graphs with $k+1$ edges. So if we show that for a graph $H$ vertices and $k+1$ the graph doesn't have a.a.s., then we immediately conclude that doesn't have any such subgraphs with $k$ vertices $k+1$ edges (Since the subgraphs are not necessarily this also rules out subgraphs with $k$ vertices more than $k+1$ edges.) As an upper bound, the expected number of labeled $H$-subgraphs in is less than $n^{|V(H)|}p^{|E(H)|}$. power of $n$ actually most with factors.) If has $k$ vertices and $k+1$ edges, this is $n^k p^{k+1}$. We $p \ll n^{-1+\epsilon}$ for all $\epsilon>0$, so in particular, $p + Therefore $p^{k+1} \ll and $n^k p^{k+1} \to as \to \infty
 n$, and this inequality is potenTially stroNg enoUgh To pRoVe thAt foR any given $S$, we haVE $|E_S| \lE |S|$ with high probability. BUt thaT's NOt whAT wE want: We want a REsULT thAt HoLds FoR AlL $S$. 
ConSidEr $k=4$, for eXample: then TheRe Are $\binom n4$ setS $s$ wE could consIdeR, and if the proBabIlity iS $\lL \frAC{4}{n^{1-\epsIloN}}$ for eAch one, THat onlY tells us tHaT The expECted numBER oF setS $S$ with $5$ edges is $\ll \biNOm N4 \Frac{4}{n^{1-\epsilon}}$ or In otheR wORdS IT is $\Ll n^{3+\Epsilon}$, whiCh Still COuld be vERy LARGe.

WE Can improve on MArkov's inequALitY for boUnDinG The proBabilItY ThaT a set $S$ inducEs a sUbgraph we Don't liKE.
First oF All: for aNy $k$, theRe iS onLy a cONsTaNt nUmBEr oF $K$-vErtEX grAphs with $K+1$ eDgEs. So iF we sHOW THat fOr a FixeD grapH $H$ (on $k$ vertices And With $K+1$ EdgEs) the RandoM graPh Doesn'T have aN $H$-subGrAph a.a.s., then we immEdiaTely conclUde ThAt tHe RandoM Graph dOesN't hAve any sUch subgRAphS wITH $K$ vErtices and $k+1$ edges a.a.S. (SINCe The subgrAphs arE NoT nECessarilY iNduCed, tHIS also RuleS OuT subgrapHs with $K$ VeRtIces and MoRe than $K+1$ eDgeS.)
As An uppER bouNd, the eXpected nUmber OF labeled $H$-subgrAPhs in $\mathcal G_{N,P}$ iS LEsS Than $N^{|V(H)|}P^{|E(H)|}$. (The power Of $n$ iS ActuAlly AT mOst $N(N-1)(n-2)\dotSb$ witH $|V(h)|$ FaCTors.) If $H$ has $k$ vertices AnD $k+1$ edgeS, this Is $n^k p^{k+1}$. 
We have $p \Ll n^{-1+\epsilon}$ FOR All $\epsilOn>0$, so IN pARticular, $p \ll n^{-1 + \frAc{1}{k+1}}$. ThErefore $p^{k+1} \lL N^{-k}$, and $n^k p^{K+1} \to 0$ as $N \to \infty
 n$, and this inequality i s potentia lly s tro ngen ough  toprove that for anygiven $S$, we have $|E _S| \ le |S|$ wi th hi gh prob a bi l i ty.  B ut  th at ' snot w hat  we wan t: we want  are sult that ho l ds  for all $ S$.  
Consider $ k=4 $, for  e xam p le: t hen  ther e are$ \binom  n4$ sets  $ S $ we c o uld con s i de r, a nd if the probabi l it y  is $\ll \frac {4}{n^ {1 - \e p s ilo n}} $ for each  o ne, t h at only te l l s  us that the expe cted number ofsets $ S$  wi t h $5$edges  i s  $\ ll \binom n 4 \f rac{4}{n^ {1-\ep s ilon}}$ or in o ther w ord s i t is $\ ll  n^ {3 + \ep s il on} $ , w hich sti ll  c ouldbe v e r y larg e.
Wecan i mprove on Mar kov 's i n equ ality  forboun di ng th e prob abili ty  that a set $S$  ind uces a su bgr ap h w edon't like.Fir stof all:  for an y  $k $, t h er e is only a consta nt n um ber of $ k$-ver t ex  g r aphs wit h$k+ 1$ e d g es. S o if we  show th at for afi xed gra ph  $H$ ( on  $k $ v ertic e s an d with  $k+1$ e dges) the random gra p h doesn't hav e  a n $H $ -sub gra ph a.a.s.,then we i mmed i at ely concl ude t ha t  t h e random graph does n' t have  anysuch subgraph s with $k$ v e rtices a nd $ k +1 $  edges a.a.s.(Sinc e the subg r aphs are  notnecessar ily induc e d , this a lso  ru les  ou t su bgraphs with$ k $ ve rt ices an d m ore tha n $ k+1 $ e dge s. )
As an u pper bou nd ,th eexp ected number o flab el ed$H$-s u bgraph s in$\ma th ca l  G_ {n,p}$i sl e ss t ha n$n^{ |V( H) |}p^{ |E(H ) |}$ . (Thepower of$n$ is a ct ua lly atmost $n(n-1)( n- 2)\dotsb$wi th$|V(H) | $  factors .) If $H$ has $k$ verti c es and$k+ 1$ ed ges,  this is$n^ k p^{k +1} $ . 
Wehave $ p \ll  n ^{- 1 + \epsi l o n} $ f or  all $\eps i l on> 0$, s oin p articul ar, $p \ll n^{-1 + \fr ac{1}{k+1}}$.  Th eref o r e$p^ { k+ 1 } \ ll n^{ - k }$, and $n^k p^ {k+1} \to0$ as  $n \to \i n fty 
 n$,_and this_inequality is potentially strong_enough to_prove_that for_any_given $S$, we_have $|E_S| \le_|S|$ with high probability._But that's not_what_we want: we want a result that holds for all $S$. 
Consider $k=4$, for_example:_then there_are_$\binom_n4$ sets $S$ we could_consider, and if the probability_is $\ll_\frac{4}{n^{1-\epsilon}}$ for each one, that only tells us_that_the expected number_of sets $S$ with $5$ edges is $\ll \binom_n4 \frac{4}{n^{1-\epsilon}}$ or in other words_it is $\ll_n^{3+\epsilon}$,_which_still could be very_large.

We can improve on Markov's inequality_for bounding the probability that a_set $S$ induces a subgraph we don't_like.
First of all: for any $k$,_there is only a constant_number of_$k$-vertex graphs with $k+1$ edges._So if we_show that_for a fixed_graph $H$ (on $k$ vertices and_with $k+1$ edges)_the random graph doesn't have an_$H$-subgraph_a.a.s., then we_immediately_conclude_that the_random graph doesn't_have_any such_subgraphs_with $k$ vertices and $k+1$ edges_a.a.s._(Since the subgraphs are not necessarily induced,_this also rules out_subgraphs_with $k$ vertices and_more than $k+1$ edges.)
As an_upper bound, the expected number of_labeled $H$-subgraphs_in $\mathcal_G_{n,p}$ is less than $n^{|V(H)|}p^{|E(H)|}$. (The power of $n$ is actually_at most $n(n-1)(n-2)\dotsb$ with $|V(H)|$ factors.)_If $H$ has $k$_vertices and_$k+1$_edges, this is_$n^k_p^{k+1}$. 
We_have $p \ll n^{-1+\epsilon}$ for all $\epsilon>0$,_so in_particular, $p \ll n^{-1 + \frac{1}{k+1}}$._Therefore $p^{k+1} \ll n^{-k}$,_and_$n^k p^{k+1} \to 0$ as $n_\to \infty