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 $x> s$. 

Very good. To finish off, you may want to say something like: "since $x\leq s+\epsilon$ for all $\epsilon>0$, it follows that $x\leq s$, a contradiction".
