Prove Halting on all Inputs is not in RE simulation
What is the Halting on all inputs problem?

It is the problem about the language $TOTAL=\{⟨M⟩\mid M \text{ is a TM and }M\text{ halts on all inputs} \}$.

What are all inputs? Let us fix $\Sigma$, the set of input symbols for Turing machines. The set of all inputs is $\Sigma^*$, the Kleene star of $\Sigma$. Let $E$ be a Turing machine that computes a one-one correspondence $cor$ from $\Sigma^*$ to nonnegative numbers.



In order to specify the behavior of a Turing machine $M$ whose language/encoding is in $TOTOAL$, we will just specify its behavior when the input is a nonnegative integer $n$ with the understanding that we are actually specify its behavior on the input that corresponds to $n$ under $E$.

What is the complement of the halting Halting problem?

It is the problem about the language $\overline{HALT}= \{⟨M, w⟩\mid M \text{ is a TM and }M\text{ does not halt on input } w\}$. 

Given a word $⟨M, w⟩$ where $M$ is a Turing machine and $w$ is a input word, how can we construct a new Turing machine related to the way how $TOTAL$ is specified?


  $M'$ = simulate $M(w)$ for $n$ steps; if $M(w)$ halts, make it loop else, stop.


The above quotation of a construction is too brief to be understood correctly easily. Here is a more detailed construction that should be easier to understand.


  $M' = $ "On input $x$, use $E$ to compute $cor(x)$. Simulate $M$ with input $w$ as long as possible but for no more than $cor(x)$ steps. If the simulation halts, continue to loop forever; otherwise, halt."


Now you should be able to check that 
$$⟨M, w⟩\in\overline{HALT} \Longleftrightarrow M'\in TOTAL.$$ 

The too-brief version equates $x$ with $cor(x)=n$, which could be very confusing for beginners although acceptable for users experienced in the theory of computability. 


  I don't understand why do I have to make M' simulate M(w) for N STEPS. Why can't I just simulate it one and that's it?


The "N" in "N STEPS" is/represents/encodes the input to $M'$. It is a variable independent of $M'$.