Every compact subspace in a metric space is bounded and closed I'm trying to prove this proposition by supposing that there exists a compact subspace S of (X, $\tau$) such that $S$ is not bounded.
As every metric space is a Hausdorff space and every compact subspace of a Hausdorff space is closed we get the first part.
Now, as to prove that $S$ is bounded, I've tried the following way:
Lets suppose that S is not bounded and lets see that this creates a contradiction.
Given $s_0 \in S$ we see that $\exists$ $\mathscr{U}$ an open cover such that $\mathscr{U}$=$\cup _{s \in S}$B($s_0$ ; $d$($s_0$, $s$)+$\epsilon$) now as S is compact $\exists$ $\mathscr{U}_f$ a finite cover of S such that $\mathscr{U}_f$=$\cup _{i \leq n}$B($s_0$ ; $d$($s_0$, $s_i$)+$\epsilon$) with $n \in \mathbb{N}$. However, as $S$ is not bounded, given M = 2$\cdot$max{$d(s_0, s_i)+\epsilon$}$_{i\leq n}$ $\exists s_1, s_2 \in S$ such that $d(s_1, s_2)>$M therefore at least one $s_1$ or $s_2$ isn't and element of $\mathscr{U}_f$ contradicting that $\mathscr{U}_f$ is a subcover and therefore that S is compact which goes against the hypothesis.
My problem with this proof is that I don't know if its rigurous enough and I've failed to find a similar proof as they all aproach the problem directly so I'd appreciate some feedback.
Thanks in andvance.
 A: 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.
A: 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.
