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 $F = \emptyset$; 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 $X = U_{j_0} \cup \dots \cup U_{j_n}$, unless $X = \emptyset$, in which case $F = \emptyset$ 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 $F = \emptyset$, 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 $X$.