A closed subset of a compact metric space is compact

$X\setminus F$ is open in $X$.

By the ambient characterisation it suffices to show that every family $(U_i{)}_{i\in I}$ of open subsets of $X$ with $F\subseteq {\bigcup }_{i\in I}{U}_i$ has finitely many members whose union contains $F$, or that $F=\mathrm{\varnothing }$; so fix such a family.

Take an object $\ast$ not in $I$, put $I^{+}:=I\cup \{\ast \}$ and $U_{\ast }:=X\setminus F$; then $(U_i{)}_{i\in I^{+}}$ is a family of open subsets of $X$ whose union is $X$, since a point outside $F$ lies in $U_{\ast }$ and a point of $F$ lies in some $U_i$ with $i\in I$.

Applying the ambient characterisation to the compact subset $X$ of itself gives $n\in \mathbb{N}$ and $j_0,\dots ,j_n\in I^{+}$ with $X=U_{j_0}\cup \cdots \cup U_{j_n}$, unless $X=\mathrm{\varnothing }$, in which case $F=\mathrm{\varnothing }$ and there is nothing to prove.

Delete from the list $j_0,\dots ,j_n$ every entry equal to $\ast$; what remains is a finite list of indices from $I$, possibly empty, and the union of the corresponding sets still contains $F$, because $U_{\ast }=X\setminus F$ contains no point of $F$ while every point of $F$ lies in one of the listed sets.

If that remaining list is empty then $F=\mathrm{\varnothing }$, and otherwise it exhibits finitely many members of $(U_i{)}_{i\in I}$ whose union contains $F$; in both cases the condition of step 1.2 is met, so $F$ is a compact subset of $X$.