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:


*

*$cl(M) = M;$

*$cl(M) \neq M$.

 A: 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.
A: For the proof above, there is no need to do two cases.
Let $\epsilon > 0$. Since $M$ is totally bounded, there exists a finite collection of open balls $\left\{B_i\right\}_{i=1}^n$ of radius $\frac{\epsilon}{2}$ covering $M$. Each ball $B_i$ is centred at a point $x_i$.
Let $x\notin M$ be a limit point of $M$. Since $x$ is a limit point, the punctured ball of radius $\frac{\epsilon}{2}$ around $x$ contains a point $y \in M$. Furthermore, $y\in B_k$ for some $k$.
We can create a new cover $\left\{C_i\right\}_{i=1}^n$ where $C_i$ is the ball centred at $x_i$ with radius $\epsilon$. Note that $C_k$ contains $x$.
A: Note firstly that an open ball is contained in a closed ball of the same radius, and a closed ball is contained in an open ball of any larger radius.


*

*By definition, if $S$ is totally bounded it is covered by a finite number of open balls radius $\epsilon/2$ centered on points in $S$.

*Closed balls of the same radius  $\epsilon/2$ on the same centres therefore also cover $S$.

*Their finite union, $C$ is a closed set containing $S$.

*By definition, $cl(S)$, the closure of $S$ is the intersection of all closed sets containing $S$, and is a subset of every closed set that contains $S$ (in particular, $C$).

*So, $C$ contains $cl(S)$.

*Open balls of radius  $\epsilon$ on  the same centres cover the corresponding closed balls, therefore cover $C$ , therefore cover  $cl(S)$.


So, $cl(S)$ is totally bounded. 
(adapted from https://math.stackexchange.com/q/693968)
A: My answer is $6-7$ years late, but I am posting it anyway since I did not find my idea being used in any of the other solutions. Not sure what definition of totally bounded you're using, but I shall use the following characterization from Carothers' Real Analysis:

In $(M,d)$, $A\subset M$ is totally bounded iff for all $\epsilon > 0$ there are finitely many sets $A_1,A_2,\ldots,A_n\subset M$ with $\text{diam}(A_i) < \epsilon$ for $1\le i\le n$, such that $A\subset \bigcup_{i=1}^n A_i$.

In some sense, $A$ is being covered by $A_i$, $1\le i\le n$. If you're using some other definition/characterization, the equivalence should be easy to prove.
Using this, you may prove the following Lemma:

Lemma $1.1$: In $(M,d)$, $A\subset M$ is totally bounded iff for all $\epsilon > 0$ there are finitely many closed sets $A_1,A_2,\ldots,A_n\subset M$ with $\text{diam}(A_i) < \epsilon$ for $1\le i\le n$, such that $A\subset \bigcup_{i=1}^n A_i$.

This should follow easily using ideas such as $\text{diam}\ A = \text{diam}\ \overline A$, $A \subset \overline A$ and basic definitions. Once you've proved Lemma $1.1$, you're good to go.
Suppose in a metric space $(M,d)$, $A\subset M$ is totally bounded. Then, for given $\epsilon > 0$ there exist $A_1,\ldots,A_n$ such that $$A\subset\bigcup_{i=1}^n A_i$$
Using Lemma $1.1$, we may assume that all $A_i$'s are closed, and hence so is their finite union. $\overline A$ is the smallest closed set containing $A$, so it immediately follows that $$\overline A\subset\bigcup_{i=1}^n A_i$$ and we are done! We have found closed sets $A_i$ which cover $\overline A$, and Lemma $1.1$ says that we have to do no more.

At this stage, I urge you to prove the following double-implication, part of which we have already done:

In a metric space $(M,d)$, $A\subset M$ is totally bounded if and only if $\overline A$ is totally bounded.

