**Proposition**: Let $(X,\mathcal{T})$ be a [[Hausdorff topological spaces|Hausdorff topological space]] and $K \subset X$ be [[Compact topological spaces|compact]]. Then $K$ is closed. **Proof**: Let $x \in X \backslash K$. To show that $X\setminus K$ is open, it suffices to show that there exists an open neighbourhood of $x$ in $X$ which does not intersect $K$. By the assumption that $X$ is Hausdorff, for all $y \in K$ , there exist disjoint open sets $U_{y}, V_{y}\in \mathcal{T}$ containing $y$ and $x$ respectively. Clearly the union of open sets $\{ U_{y} : y\in K \}$ forms an open cover of $K$. ![[Drawing - Compact Subspaces of Hausdorff Spaces are Closed.excalidraw]] Hence by assumption that $K$ is compact, there exists a finite subcover $U_{y_{1}}, \dots, U_{y_{n}}$ of $K$. Since $\mathcal{T}$ is closed under finite intersections, the intersection $V:=\bigcap_{i = 1}^n V_{y_{i}}$is still open, and by construction it is disjoint from all $U_y$ for $y \in S$, hence in particular disjoint from $K$, and it contains $x$. Hence $V$ is an open neighbourhood of $x$ as required.