I was thinking about compactness and found the (I'm sure already well-known) property that relatively closed subsets of compact sets are compact.
The proof I found is the following:
Let $F \subseteq K \subseteq X$ where $X$ is a metric space and $K$ is a compact subset. Since $F$ is relatively closed in $K$ we have that $K \setminus F = K \cap H$ for some open $H \subseteq X$. Taking an open cover $\{G_\alpha\}$ of $F$ in $X$, we have $F \subseteq \cup_{\alpha}G_\alpha$. We also have $K = F \cup K\setminus F \subseteq (\cup_{\alpha}G_\alpha) \cup H$ which is an open cover of $K$ so by compactness there is a finite subcover. If this finite subcover includes $H$ then throw it out, if not then use it as is. This yields a finite subcover of $F$, so $F$ is compact since the original open cover was arbitrary.
Does this proof make sense? I felt it kind of odd since being open and closed is a relative property and so $F$ being closed with respect to $K$ doesn't mean it has to be closed with respect to $X$ but compactness with respect to $X$ forces $F$ to be closed.