(Theorem) Compact subspaces of Hausdorff spaces are closed

Theorem (26.3): Every compact subspace of a Hausdorff space is closed.

Proof from Modern Analysis I.

Let $X$ be Hausdorff and $K\subseteq X$ be compact. We will show that $A$ is closed by showing that the complement $X\backslash K$ is open.

Fix $x\in X\backslash K$. By the Hausdorff property, for every $y\in K$, we may find a neighborhood $U$ of $x$ and $V$ of $y$ such that $U\cap V=\varnothing$. Then ${\bigcup }_{y\in Y}{V}_y$ is a cover of $K$, and by compactness, it has a finite subcover. Then there is a corresponding finite intersection ${\bigcap }_{y\in Y}{U}_y$ of neighborhoods of $x$, which is itself an open neighborhood of $x$ disjoint from $K$. Since $x\in X\backslash K$ is arbitrary, we conclude that every point of $X\backslash K$ is an interior point.