You seem to have one of the right core idea with using pigeonhole principle, but there are serious issues with the details.
You started with the family of all $\varepsilon$-balls, and you know from compactness that you can find a finite subfamily of them that cover the space. You decided to denote the centres of this finite family by $x_1, \ldots, x_{n_\varepsilon}$, which is not permissible, since $x_1, \ldots x_{n_{\varepsilon}}$ already refers to the first $n_\varepsilon$ terms of the sequence $(x_n)$, which you fixed at the beginning of the proof. You have no reason to assume that the centres of your balls lie on the sequence points.
As I said, the pigeonhole principle idea is the right idea. If $y_1, \ldots, y_n$ are centres of open $\varepsilon$-balls that cover the space, then infinitely many terms of the sequence (as opposed to $X$, as your proof states) must enter one of the balls. This will lead to the right proof.
Then your proof goes straight to the conclusion. There is an issue with this too. You've established an infinite set of sequence points that are all within $2\varepsilon$ distance from each other. But, this doesn't necessarily imply the existence of a convergent subsequence. As it stands, you're defining a different subsequence for each $\varepsilon$, when you really need a single subsequence that works for each $\varepsilon$.
Consider, instead of using arbitrary $\varepsilon$, using $\varepsilon = \frac{1}{m}$, where $m \in \Bbb{N}$. Define your subsequence recursively. Start with $\varepsilon = \frac{1}{1}$, and by the pigeonhole principle, there must be a ball $B_1$ of radius $1$ containing infinitely many points of the sequence $(x_n)$. Also let,
$$S_1 = \{k \ge 1 : x_k \in B_1\}.$$
Note that $S_1$ is infinite, hence non-empty, and has a minimum element. We can make this minimum element $n_1$, so that $x_{n_1}$ is the first term in the subsequence.
Then, since $S_1$ is infinite, another pigeonhole principle application applied to it (as opposed to our original sequence) will yield another ball $B_2$, now of radius $\frac{1}{2}$, so that $x_k \in B_2$ for infinitely many $k \in S_1$. Let
$$S_2 = \{k > n_1 : x_k \in B_2\}$$
and let $n_2$ be the minimum of $S_2$.
Continue this process, and you'll obtain a sequence of nested sets
$$S_1 \supseteq S_2 \supseteq \ldots,$$
and their minimum elements form a subsequence $x_{n_1}, x_{n_2}, \ldots$ (if you've done it right, you should be able to justify why $n_1 < n_2 < n_3 < \ldots$, which is necessary to call it a subsequence).
Why is this subsequence convergent? And what does it converge to? Well, answering the latter question first, the closure of the sequence of open balls $(\overline{B}_m)$ form a nested sequence of compact sets, which must have a non-empty intersection. Since their diameters (which are less than or equal to $\frac{2}{m}$) converge to $0$, this intersection must be a singleton. Let $x$ be the unique point in this intersection.
Why is the subsequence converge to $x$? Well, for every $\varepsilon > 0$, find some $m$ such that $\frac{2}{m} < \varepsilon$. Then,
$$k \ge m \implies x_{n_k} \in \overline{B}_m \implies d(x, x_{n_k}) \le \frac{2}{m} < \varepsilon,$$
since $x \in \overline{B}_m$ (it's in all the $\overline{B}_i$s).
As you can see, there are some serious steps missing. The core ideas are pigeonhole principle, and Cantor's intersection theorem (guaranteeing the point of convergence).
Best Answer
I think this proof might be a little clearer in its more general setting. That is if $X$ is a Hausdorff space and $Y \subset X$ is compact then $Y$ is closed. We call a space Hausdorff if for any two distinct points $x,y \in X$ there exists open neighborhoods $U,V \subset X$ so that $x \in U$ and $y \in V$ and $U \cap V =\emptyset$. It's clear that metric spaces are Hausdorff because you can separate two points taking balls around each half the distance between the two. Concretely we take $r=d(x,y)/2$ then have $U=B(x,r)$ and $V=B(y,r)$.
Now let $Y \subset X$ be compact. Pick $y \in X \setminus Y$. For each $x \in Y$ by the Hausdorff property we can find open sets $U_x$ and $V_x$ so that $x \in U_x$, $y \in V_x$ and $U_x \cap V_x=\emptyset$. Now it follows that
$$Y \subset \bigcup_{x \in Y} U_x,$$
since $Y$ is compact there exists some $x_1,\dots,x_n$ such that $U_{x_1},\dots,U_{x_n}$ covers $Y$. Then set $V=V_{x_1} \cap \cdots \cap V_{x_n}$ it follows that $V$ is open and contains $y$ but $U_{x_i} \cap V_{x_i}=\emptyset$. Since $Y$ is contained in the union of the $U_{x_i}$ we have that
$$Y \cap V = \emptyset.$$
Thereby around every point $y \in X \setminus Y$ we can find an open set contained in $X \setminus Y$. It follows that $Y$ is closed since $X \setminus Y$ is open.