Showing that the closure of a totally bounded set is totally bounded I would like to show that if I have a subset $M$ of a metric space $(X,d)$ such that $M$ is totally bounded, then its closure $cl(M)$ is also totally bounded. 
My general strategy would be to show that for every $\epsilon > 0$, there exist $x_1,...,x_n \in M$ such that $M = \bigcup B_{\epsilon/2}(x_i)$. And then I would  use the fact that this works also for $\epsilon$ since it works for $\epsilon /2$. However, should I be having two cases? I am listing the cases as follows:


*

*$cl(M) = M;$

*$cl(M) \neq M$.


Just as you mentioned, given $\varepsilon>0$, we can find $x_1,\ldots,x_n\in M$ such that $M\subseteq \bigcup_i B_{\varepsilon/2}(x_i)$. Let's show that $\operatorname{cl}(M)\subseteq\bigcup_i B_\varepsilon(x_i)$.
Let $z\in \operatorname{cl}(M)$. Then there exists $x\in M$ such that $d(z,x)<\varepsilon/2$, and there is some $i$ such that $d(x,x_i)<\varepsilon/2$, so $d(z,x_i)<\varepsilon$. Therefore, $\operatorname{cl}(M)\subseteq\bigcup_i B_\varepsilon(x_i)$.
Therefore, $\operatorname{cl}(M)$ is totally bounded.
