Why do they use $\frac{1}{n}$ balls in this proof for closed sets? 
I can follow the proof all the way upto the fact that we need to show that no $B_\epsilon(x)$ lies completely in $T$ (thus the contradiction later).
If no ball lies completely in $T$, then it must spill over to the set $S$, hence
$\forall \epsilon >0, B_\epsilon (x) \cap S \neq \emptyset $.
But what's happening next. Is $x_n$ the $n$-th term in some sequence, contained in $S$? Suppose it is, then, are we saying this sequence converges to $x$, for $\epsilon = \frac{1}{n}$ Why? I'm very confused with the line:
"there is an element $x_n$ of $S$ in $B_{1/n}(x)$..." going forward, can someone clear this up?
 A: In such proofs, you can always use $\frac{1}{n}$ instead of $\varepsilon$, because of Archimedean property. If some property is satisfied by $\frac{1}{n}\forall n\in \mathbb N$ then choosing an $\varepsilon>0$, we can choose $\frac{1}{n}<\varepsilon$. Then, since the property is satisfied for all $\frac{1}{n}$ where $n\in\mathbb N$, it is satisfied by $\varepsilon$.
This trick has wide range of applications in probability, because it is used for discretising uncountable sets (note that countable summability is an axiom of probability).
EDIT
Let me expain line by line.
You understand the statement for $\varepsilon$, i.e. $\forall \varepsilon>0 B_\varepsilon (x)\cap S\neq \emptyset$. Since this statement holds for all positive $\varepsilon$, it particularly holds for $\frac{1}{n}\forall n\in\mathbb N$. That is what the next statement is saying.
Since, these sets are not null, in the next statement author is picking elements $x_n\in B_\frac{1}{n}(x)\cap S$. Since the distance between $x_n$ and $x$ goes to zero, $x$ is the limit the sequence of $x_n$'s.
Now, in the final statement author uses the fact that limit of sequences in closed sets, lie in closed sets.
Hope it helps:)
