Trying to complete a proof for my real analysis course (total boundedness) I'm trying to prove that any totally bounded subset $S$ of a metric space $X$ contains finitely many points such that the union of the open epsilon balls centered at these points includes the set $S$.

  
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*Given that $S$ is a subset of a metric space $X$, prove that $S$ is totally bounded if and only if for every $\epsilon>0$ it is possible to find finitely many points $x_1,x_2,\ldots,x_n$ that belong to the set $S$ such that $$S=\subseteq\bigcup_{j=1}^nB\left(B_j,\epsilon\right)$$
  

Proof:
First we will assume that the set $S$ is totally bounded and prove that for every $\epsilon>0$ it is possible to find finitely many points $x_1,x_2,\ldots,x_n$ that belong to the set $S$ such that $$S=\subseteq\bigcup_{j=1}^nB\left(B_j,\epsilon\right)$$
Suppose that $\epsilon>0$.
If $S$ is empty, then any union of balls entered at any point in $S$ will be empty. Since $0\subseteq0$, our condition is met.
The case where $S$ is empty seems awkward, should I have included this?
If $S$ is nonempty, we can choose a point $x_1\in S$.
The set $S$ is either included in the ball $B\left(x_1,\epsilon\right)$ or $S$ is not included in $B\left(x_1,\epsilon\right)$.
In the case that $S\subseteq B\left(x_1,\epsilon\right)$, we have satisfied our conclusion.
In the case that $S\nsubseteq B\left(x_1,\epsilon\right)$, we may choose a different point $x_2\in S$ such that $x\notin B\left(x_1,\epsilon\right)$.
Note again that we have either $S\subseteq\left(B\left(x_1,\epsilon\right)\cup B\left(x_2,\epsilon\right)\right)$ or $S\nsubseteq\left(B\left(x_1,\epsilon\right)\cup B\left(x_1,\epsilon\right)\right)$.
If $S\nsubseteq\left(B\left(x_1,\epsilon\right)\cup B\left(x_1,\epsilon\right)\right)$, we may choose a point $x_3$ in $S$ such that $x_3\nsubseteq\left(B\left(x_1,\epsilon\right)\cup B\left(x_1,\epsilon\right)\right)$.
I want to say that if we continue on in this manner we will eventually find some $n$ such that $$S=\subseteq\bigcup_{j=1}^nB\left(B_j,\epsilon\right)$$ but I don't know how to state that mathematically. I know that I haven't used the fact that $S$ is totally bounded, and I feel like there is some fundamental flaw with what I'm trying to do.
I'm stuck with trying to continue a process without actually continuing the process… Is it okay to just say something like "continue this process and eventually you will find an $n$ such that this is true"? I feel like it's not.
 A: First of all, it isn't necessary to treat the empty set specially, a general argument will suffice. In response to your next query, why must the process terminate? If you can think of a mathematical reason for it to terminate (and yes, here you would have to use total boundedness), then you can state that and claim the process terminates.
However, the forward direction you're attempting to show can be seen in much simpler manner: think about the definition of total boundedness!
If a space $X$ is totally bounded, then for every $\epsilon > 0$, there exists a finite cover of $X$ by epsilon balls, i.e. for some $\left\{B(x_i;\epsilon)\right\}_{i = 1}^n$,
$$
X\subseteq\bigcup_{i = 1}^n B(x_i;\epsilon),
$$
where $x_i\in X$ for all $i$. If you look at this cover, you should be able to find your finite set of points for a fixed $\epsilon$ and complete your proof.
The reverse direction should be similar: follow the definitions!
EDIT (in response to revision of proof): I have to ask for your definition of totally bounded. In the definition I'm familiar with, you choose an $\epsilon > 0$, and this gives you an open cover of $X$ by finitely many $\epsilon$-balls - you don't choose $n$ and find a covering of $\epsilon$-balls that has $n$ balls.
Second, I have an issue with your claim that choosing an arbitrary element $x_i\in S\cap B(y_i;\epsilon)$ and taking the balls $B(x_i;\epsilon)$ will cover $S$: let $S = [0,1]$ (with the Euclidean metric). This is totally bounded, so we can take $\epsilon = .5$ and notice that the balls $B(1/4;\epsilon)$ and $B(3/4;\epsilon)$ cover $S$. However, take your $x_1 = 0$ and $x_2 = 1$: then $S\not\subseteq B(x_1;\epsilon)\cup B(x_2;\epsilon)$.
Finally, it seems to me that when you try to prove the reverse direction, you wind up assuming what you want to show: namely, that $S$ is totally bounded ("using the fact that $S$ is totally bounded...").
So: what is your definition of totally bounded, first of all? Secondly, I think you're looking a little too hard for your finite set of $x_i$'s. Take a step back, and look at the definition of totally bounded: there should be a natural choice for the $x_i$'s. Further, it should be enough to show that this holds when you have a metric space that is itself totally bounded (can you see why?). I think that the notion of the subset being totally bounded rather than just the space being totally bounded might also be hampering your understanding and insight into how you might solve the problem (and if not, it's still a good idea to think about why this is enough). In any case, keep at it! It's nice to see someone actively trying to solve their own problem on here rather than expecting a solution.
Edited again: Seeing your definition, there isn't as nice of a choice for the $x_i$'s as I had thought. However, I have some other tips on getting a solution. First, you know that you can find a finite cover of $\epsilon$-balls for any $\epsilon > 0$. So why not start with a cover by smaller balls than you need? That is, take a cover by $\epsilon/2$-balls, then increase the radius and work with it. You'll have more balls with which you can cover $S$. Another thought: in a metric space, we have a notion of the distance between a point and a set. Perhaps you can use this notion to help make your estimates more precise.
