Proof that compact subsets of a Hausdorff space are closed

Proof

We show that the complement of $K$ is open, by separating each outside point from $K$.

1. Let $X$ be a Hausdorff space and let $K$ be a compact subset of $X$.
2. If $K = \emptyset$, then $X \setminus K = X$ is open, and $K$ is closed. So assume $K \ne \emptyset$.
3. Fix $y \in X \setminus K$. For each $x \in K$ we have $x \ne y$. By the Hausdorff property, choose open sets $U_x \ni x$ and $V_x \ni y$ with $U_x \cap V_x = \emptyset$.
4. The sets $U_x$ with $x \in K$ are open and cover $K$. By compactness via open covers, there are $x_1, \ldots, x_n \in K$ with $K \subseteq U_{x_1} \cup \cdots \cup U_{x_n}$, and $n \ge 1$.
5. Let $V_y = V_{x_1} \cap \cdots \cap V_{x_n}$. A topology is closed under finite intersections, so $V_y$ is open. Also $y \in V_y$.
6. Let $z \in K$. By step 4, $z \in U_{x_j}$ for some $j$. Since $U_{x_j} \cap V_{x_j} = \emptyset$ and $V_y \subseteq V_{x_j}$, we get $z \notin V_y$.
7. So $V_y \subseteq X \setminus K$ for each $y \in X \setminus K$.
8. Hence $X \setminus K = \bigcup_{y \in X \setminus K} V_y$, a union of open sets. It is open by the axioms of a topology.
9. Thus $K$ is closed in $X$.