Every compact subspace in a metric space is bounded and closed

Your proof has the right ideas, albeit could be a little neater, and there are some notational issues. E.g. you write $\mathscr{U} = \cup_{s \in S} \: B(s_0; d(s_0,s)+\varepsilon)$, but remember the cover $\mathscr{U}$ is the collection of subsets whose union contains $S$, not the union itself.

I would make the following modifications if you want to use your argument. Suppose $S$ is closed and nonempty, and fix $s_0 \in S$. Let $\varepsilon > 0$. Consider the open sets $U_s = B(s_0, d(s_0,s)+\varepsilon)$, defined for each $s \in S$. Evidently, $\{U_s\}_{s\in S}$ is an open cover of $S$. By assumption, $S$ is compact, so there exist $s_1,\dots,s_n \in S$ such that $S \subseteq U_{s_1} \cup \dots \cup U_{s_n}$. In particular, if $M = \max\{d(s_0,s_j)\; | \; 1 \leq j \leq n\}$, then for any $s \in S$, we have $d(s_0,s) \leq M + \varepsilon$. Hence $S$ is bounded by triangle inequality.

You can prove boundedness directly. Arbitrarily choose any $y \in S$. Then for $n \in \Bbb N$ let $U_n= \{x \in S \mid d(x, y) \lt n \}$. Then $\{U_n \mid n \in \Bbb N \}$ is an open cover of $S$. Since $S$ is compact, this open cover must have a finite subcover. But $m \lt n \Rightarrow U_m \subseteq U_n$, so a finite subcover is simply $U_N$ for some $N \in \Bbb N$. Thus $S \subseteq U_N$, so $x \in S \Rightarrow d(x, y) \lt N$ and $S$ is bounded.

You can use a similar idea to directly prove that $S$ is closed. If $x \notin S$, for $n \in \Bbb N$ let $U_n = \{y \in S \mid d(y, x) \gt \frac 1n \}$. Then again, the $U_n$ are a nested open cover of $S$, so a finite subcover must be a single set $U_N$ for some $N \in \Bbb N$. But then $B \left (x, \frac{1}{N+1} \right ) \cap S = \varnothing$ so $x \notin \overline S$. Since $x$ is an arbitrary point outside of $S$ , it follows that $\overline S \subseteq S$. But it's always the case that $S \subseteq \overline S$, so equality follows.