If a set is closed, why is that set intersected with a compact set closed?

Question

If $F$ is a closed subset of $K$ and $K$ is compact, why is $F \cap K$ closed?

Progress

I just realized compact subsets of a metric space are closed.

Answer 1

For completeness, it might be worth explaining why a compact subset of a metric space (or more generally, a Hausdorff space) is closed.

Suppose $X$ is a Hausdorff space and $K$ is a compact subset of $X$. Let $x \in X \setminus K$. Then for every $k \in K$, we can find a pair of disjoint open sets $U_k$ and $V_k$ such that $k \in U_k$ and $x \in V_k$. (We can do this because $X$ is Hausdorff.)

Now $\bigcup_{k \in K}U_k$ is an open cover of $K$, and $K$ is compact, so there is a finite subcover, say $U_{k_1}, \ldots, U_{k_n}$. Then $V_{k_1} \cap \cdots \cap V_{k_n}$ is an open set containing $x$ and disjoint from $K$. This shows that $X \setminus K$ is open, and therefore $K$ is closed.

Now, as suggested in the comments, the desired result follows from the fact that the intersection of two closed sets is closed.