A simple(?) funny combinatorial problem!
The conjecture in the post does not hold, but the weaker conjecture (with respect to the floor) mentioned in the comments does hold.  In fact, something stronger holds.



Lemma 1. The conjecture in the post does not hold.  That is, there is an instance satisfying the given assumptions where
$$\big|\textstyle\big\{ S \subseteq [n] : \sum_{i \in S}~ c_i \geq E \, t \big\}\big| &lt; \binom{ E \, n}{ t }.$$

Proof.
Consider the instance with $n=3$, $c=(1, 1, 0.7)$, $E=2.7/3=0.9$, and $t=2$.  Then $E\, t = 1.8$.
For the left-hand side, we have
$$\big|\textstyle\{ S \subseteq [3] : \sum_{i \in S}~ c_i \geq 1.8 \}\big| = 2$$
because any subset $S$ that does not contain both 1's sums to at most 1.7, and there are only two subsets ($\{1,1\}$ and $\{1,1,0.7\}$) containing both 1's. And the right-hand side is $\binom{ 2.7}{ 2 } = 2.7\cdot 1.7\,/\, 2 = 2.295&gt;2. ~~~\Box$



The weaker conjecture suggested in the comments, namely the bound w.r.t. the floor, $\lfloor E\, n \rfloor$, does hold.  In fact something slightly stronger holds:

Lemma 2. Fix $0&lt;E&lt;1$, integers $n,t&gt;0$,
and vector $c \in [0,1]^n$ with $\sum_{i\in [n]} c_i \geq E\, n$.
Then
$$\big|\textstyle\big\{ S \subseteq [n] : \sum_{i \in S}~ c_i \geq E\, t \big\}\big| \,&gt;\, \binom{ \lfloor E\, n\rfloor}{ t } + \binom{ \lfloor E\, n\rfloor}{ t+1 } + \cdots + \binom{ \lfloor E\, n\rfloor}{ \lfloor E\, n\rfloor }.$$

Proof. Let $a=\lfloor E\, n\rfloor$. Assume WLOG that $a=E\,n$.  (Otherwise scale $E$ and each $c_i$ down by a uniform factor to make it so.  This maintains $\sum_i c_i \ge E\,n$ and changes neither which subsets sum to at least $E\,t$ nor the desired lower bound on the number of such subsets.)  Assume WLOG that $t\le a$ (otherwise the claim holds trivially).

Consider any subset $S\subseteq [n]$ of size at least $n-d$, where $d=a-\lceil a\,t/n\rceil \ge 0$. Since $\sum_{i\in[n]} c_i \ge a$ and $S$ contains all but at most $d$ elements (each of which is at most 1), we have $\sum_{i\in S} c_i \ge a-d = \lceil a\,t/n\rceil = \lceil E\,t\rceil \ge E\,t$, as desired.

The number of such subsets $S$ is 

${n\choose n-d} + {n \choose n-d+1} + \cdots + {n \choose n-1} + {n \choose n}$

$ = {n\choose d} + {n \choose d-1} + \cdots + {n \choose 1} + {n \choose 0}$

$ &gt; {a\choose d} + {a \choose d-1} + \cdots + {a \choose 1} + {a \choose 0}~~~$ (using $n&gt;a$)

$ = {a\choose a-d} + {a \choose a-d+1} + \cdots + {a \choose a-1} + {a \choose a}.$

But $a-d = \lceil a\,t/n\rceil \le t$ (using $a/n = E &lt; 1$), so the last sum is at least the desired lower bound on the number of good subsets. $~~\Box$