Closed subsets of compact sets are compact

Statement

Let $X$ be a topological space, let $K \subseteq X$ be compact and let $F \subseteq X$ be closed with $F \subseteq K$. Then $F$ is compact.

Remarks

Proof: let $(U_i)$ be open sets of $X$ that cover $F$. Add the open set $X \setminus F$ to get an open cover of $K$. A finite subcover of $K$ gives, after you remove $X \setminus F$, a finite subcover of $F$. More generally, if $K$ is compact and $F$ is closed, then $K \cap F$ is compact. For example, the Cantor set is compact, because it is a closed subset of $[0, 1]$.

Explanations

L1 human

A closed part of a compact set is again compact. The open complement of the closed part covers everything outside it, so it does the job of all the extra open sets.