Let $X$ be a Hausdorff space, and $C \subseteq X$ a compact subset. We are to show that $C$ is closed. We will do so, by showing that the complement $U = X \setminus C$ is open. To prove that $U$ is open, it suffices to demonstrate that, for each $x \in U$ , there exists an open set $V$ with $x \in V$ and $V \subseteq U$ .

Fix $x \in U$ . For each $y \in C$ , using the Hausdorff assumption, choose disjoint open sets $A_y$ and $B_y$ with $x \in A_y$ and $y \in B_y$ .

Since every $y \in C$ is an element of $B_y$ , the collection $\{B_y \mid y \in C\}$ is an open covering of $C$ . Since $C$ is compact, this open cover admits a finite subcover. So choose $y_1, \ldots, y_n \in C$ such that $C \subseteq B_{y_1} \cup \cdots \cup B_{y_n}$ .

Notice that $A_{y_1} \cap \cdots \cap A_{y_n}$ , being a finite intersection of open sets, is open, and contains $x$ . Call this neighborhood of $x$ by the name $V$ . All we need to do is show that $V \subseteq U$ .

For any point $z \in C$ , we have $z \in B_{y_1} \cup \cdots \cup B_{y_n}$ , and therefore $z \in B_{y_k}$ for some $k$ . Since $A_{y_k}$ and $B_{y_k}$ are disjoint, $z \notin A_{y_k}$ , and therefore $z \notin A_{y_1} \cap \cdots \cap A_{y_n} = V$ . Thus $C$ is disjoint from $V$ , and $V$ is contained in $U$ .