X regular and K compact prove $\overline{K}$ is compact

Question

I tried: Let $x_i\in K$ such $x_i\in U$ open. If $X$ is regular exists $V_i$ such $x_i\in K\subset V\subset\overline{V}\subset U$. If $K$ compact and a cover such $K\subset\bigcup_{x_i\in K} V_i$ exist a finite cover such $K\subset\bigcup V_{i_n}=V$. Clearly $\overline{K}\subset\overline{V}$ and $\overline{V}$ is a finite cover of $\overline{K}$

Answer 1

Your proof is incorrect, because you only showed that there is some finite (not even open, because $\overline{V}$ typically is not open) cover of $\overline{K}$. For example you can take $U=X$ and $V=X$, and you obtain that your cover is simply $\{X\}$. But you were supposed to show that every open cover has a finite subcover.

Consider an open cover $\mathcal{U}$ of $\overline{K}$. In particular $\mathcal{U}$ is an open cover of $K$ and since $K$ is compact, then there is a finite subcover $\mathcal{U}'=\{U_1,\ldots,U_n\}\subseteq\mathcal{U}$ covering $K$. We will show that $\mathcal{U}'$ also covers $\overline{K}$.

So take $v\in\overline{K}$ and assume that $v$ does not belong to any of $U_i$. Now for any $x\in U_i$ there is an open subset $V_{i,x}$ such that $x\in V_{i,x}$ and $\overline{V_{i,x}}\subseteq U_i$. That's because $X$ is regular. Now the collection

$$\big\{V_{i,x}\ \big|\ i=1,\ldots,n,\ x\in U_i\big\}$$

is again an open cover of $K$ and so it has a finite subcover, say $\{V_1,\ldots,V_m\}$. Of course $C=\bigcup_{j=1}^m\overline{V_j}$ is closed, $K\subseteq C$ and $v\not\in C$. Since $X$ is regular then $v$ and $C$ can be separated by open subsets. In particular there is an open neighbourhood $U'$ of $v$ such that $U'\cap C=\emptyset$. And so $U'\cap K=\emptyset$ and thus $v\not\in\overline{K}$. Contradiction. $\Box$