Compact subset in a Hausdorff space, is closed.

Question

Let $(X, \tau)$ be a topological Hausdorff space and $K\subseteq X$ compact.

If $(X,\tau)$ is a Hausdorff space, then is $K$ closed.

For my proof I want to show, that $K=K\cup\partial K=\overline{K}$. $K\subseteq\overline{K}$ is trivial, so I need to show $K\supseteq K\cup\partial K$:

Let $\overline{x}\in K\cup\partial K$. If $\overline{x}\in K$, there is nothing to show. Hence, suppose $\overline{x}\in\partial K$. Then we have for every neighborhood $U$ of $\overline{x}$ that $$U\cap K\neq\emptyset$$ and $$U\cap (X\setminus K)\neq\emptyset.$$

Since $\overline{x}\notin K$, $\overline{x}\in U\cap (X\setminus K)$

For $x\in K$ let $U_x$ be a neighborhood of $x$. Then $$K\subseteq \bigcup_{x\in K} U_x.$$ Since $K$ is compact there exist $x_0,\dotso, x_n$ such that $K\subseteq\bigcup_{i=0}^n U_{x_i}$ open.

Thus $$U\cap\bigcup_{i=0}^n U_{x_i}\neq\emptyset$$ and hence there exists $x_l\neq\overline{x}$ such that $U_{x_l}\cap U\neq\emptyset$ for every neighborhood $U$ of $\overline{x}$, which contradicts to $X$ beeing a Hausdorff space. Therefore, $\overline{x}\in K$ and $K$ is indeed closed.

I appreciate your thoughts on my proof.
Thanks in advance.

Answer 1

To show $K^c$ is open: Let $x\in K^c$. For each $y\in K$, let $U_y$ and $V_y$ be such that $x\in U_y, y\in V_y$ and $U_y\cap V_y=\emptyset$, by Hausdorffness. Now $\{V_y\}_{y\in K}$ is an open cover of $K$. Take a finite subcover, $V_{y_1},\dots, V_{y_n}$, by compactness. Then consider $U=U_{y_1}\cap U_{y_2} \dots \cap U_{y_n}$. It is easy to see that $U$ is a nbhd of $x$ with $U\subset K^c$. $\square $