compact subspace of a Hausdorff space is closed

Let $X$ be a Hausdorff space, and $Y$ be a compact subspace of $X$. We prove that $X\setminus Y$ is open, by finding for every point $x\in X\setminus Y$ a neighborhood $U_x$ disjoint from $Y$.

Let $y\in Y$. $x\ne y$, so by the definition of a Hausdorff space, there exist open neighborhoods $U_x^{(y)}$ of $x$ and $V_x^{(y)}$ of $y$ such that $U_x^{(y)}\cap V_x^{(y)}=\mathrm{\varnothing }$. Clearly

$$Y\subseteq \bigcup _{y\in Y}{V}_x^{(y)}$$

but since $Y$ is compact, we can select from these a finite subcover of $Y$

$$Y\subseteq V_x^{(y_1)}\cup \mathrm{\cdots }\cup V_x^{(y_n)}$$

Now for every $y\in Y$ there exists $k\in 1\mathrm{\dots }n$ such that $y\in V_x^{(y_k)}$. Since $U_x^{(y_k)}$ and $V_x^{(y_k)}$ are disjoint, $y\notin U_x^{(y_k)}$, therefore neither is it in the intersection

$$U_x=\bigcap _{j=1}^n{U}_x^{(y_j)}$$

A finite intersection of open sets is open, hence $U_x$ is a neighborhood of $x$ disjoint from $Y$.