Proposition 4.16.3. Let $X$ be a Hausdorff space and $E \subset X$ be compact, then $E$ is closed.
Proof. Let $x \in E^{c}$. For each $y \in E$, there exists $U_{y} \in \cn(y)$ and $V_{y} \in \cn(x)$ such that $U_{y} \cap V_{y} = \emptyset$. By compactness, there exists $E_{0} \subset E$ finite such that $\bigcup_{y \in E_0}U_{y} \supset E$. In which case, $E \cap \bigcap_{y \in E_0}V_{y} = \emptyset$ and $\bigcap_{y \in E_0}V_{y} \in \cn(x)$. Thus $E^{c}$ is open.$\square$