Compact subsets of a Hausdorff space are closed

Statement

Let $X$ be a Hausdorff space and let $K$ be a compact subset of $X$. Then $K$ is closed in $X$.

Remarks

Proof: fix $y \notin K$. For each $x \in K$, choose disjoint open sets $U_x \ni x$ and $V_x \ni y$. Finitely many $U_{x_1}, \ldots, U_{x_n}$ cover $K$. Then $V_{x_1} \cap \cdots \cap V_{x_n}$ is an open set that contains $y$ and misses $K$. The Hausdorff condition is needed. With the indiscrete topology on a set with two points, each one-point set is compact but not closed. Each metric space is Hausdorff, so compact subsets of metric spaces are closed.

Explanations

L1 human

In a Hausdorff space, you can separate a point outside $K$ from each point of $K$ by open sets. Compactness lets you do this with finitely many open sets at once, which gives an open set about the point that misses $K$.