Proof if $M$ is bounded, then so is its closure I want to proof that if $M\subseteq \mathbb R$ is bounded, then so is $\overline{M}$ or more precise that if $s$ is the supremum of $M$, then it is the supremum of $\overline{M}$. I came up with a proof but I am not sure if it is correct:
Let $s := \sup M$. If $M$ is closed we are done, so suppose $M$ is not closed. Suppose $s$ was not the supremum of $\overline{M}$. Then $\exists x\in \overline{M}$ with $x>s$.   
By the definition of the closure there exists a sequence $(a_n)_{n\in \mathbb N_0}$ in $M$ with $\lim_{n\to \infty} a_n = x$, i.e. for all $\epsilon> 0, |a_n -x| < \epsilon$ for all big enough $n\in \mathbb N_0$.   
But that means $$-\epsilon < a_n -x < \epsilon \Longleftrightarrow x < a_n + \epsilon \leq s+\epsilon$$
This is a contradiction to $x> s$. 

I think you show that $s$ is an upper bound of $\overline{M}$, but you don't actually show that it's the least upper bound of $\overline{M}$.
It should, however, be straightforward to show that for any upper bound $z$ of $\overline M$, $s \le z$. This should follow from the fact that any such $z$ will also be an upper bound of $M$.
You also might want to think carefully about whether you really needed a proof by contradiction here. (See, for example, Tim Gowers' thoughts on it here.) I don't think you do, and dropping the attempt entirely but maintaining the rest of the argument still works.
Consider this reframing of your proof (with some details you've supplied omitted):

Suppose $M$ is not closed, and consider $x \in \overline{M} \cap M^c$.
  There must then exist a sequence $\{ a_n \} \subset M$ such that $a_n \to x$. This means that for any $\varepsilon > 0$, $x < s + \varepsilon$, which implies that $x\le s$. $x$ was arbitrary, so $s$
  is an upper bound of $\overline M$.

