Closure of a compact set in Regular space $ X $ I am trying to show that closure of a compact set in a Regular space is compact, but I am hitting some hurdles as a compact set in a regular space need not be closed. 
This is what I started off with: If $ C $ is any compact subset,
let $ U $ be any open neighbourhood containing $ C $. Then by regularity of $ X $, $ \forall x \in C, \exists V_x : \overset-V_x \subseteq U $ where $ V_x $ is a neighbourhood of $ x $.
So $ \lbrace  V_x:x \in C\rbrace $ is an open cover for C, which by compactness of C , simplifies to $ \lbrace  V_{x_i}:x_i \in C\,\,for\,i=1,...,n\rbrace $. Now I end up looking at $ \cup \overset-V_{x_i} $ which  should contain $ \overset-C $. But I get stuck here. Should I look at the intersection rather??
 A: Hint: Suppose that $\mathcal{U} = \{ U_i : i \in I \}$ is any cover of $\overline{C}$ by open subsets of $X$.  For each $i \in I$ and $x \in U_i$ by regularity fix an open set $V_{i,x}$ such that $x \in V_{i,x} \subseteq \overline{V_{i,x}} \subseteq U_i$.  Note that $$\mathcal{V} = \{ V_{i,x} : i \in I, x \in U_i \}$$ is also a cover of $\overline{C}$ by open subsets of $X$.  More importantly, $\mathcal{V}$ covers the compact set $C$.
A: This following holds in a regular space $X$.
Lemma: Every open neighborhood $U$ of a compact set $K$ contains the closure $\overline K$ as well as a closed neighborhood of $K$.
Proof: Since $X$ is regular, every $x\in K$ has an open neighborhood $V_x$ such that $\overline {V_x}\subseteq U$. The family $\{V_x\mid x\in K\}$ is an open cover of $K$ which by compactness has a finite subcover $\{V_i,\dots,V_n\}$. Since closure commutes with finite unions, we know that $\bigcup_{i=1}^n V_i$ is a neighborhood of $K$ whose closure $\overline{\bigcup_{i=1}^n V_i}=\bigcup_{i=1}^n\overline {V_i}$ is a subset of $U$ and contains $\overline K$.
Corollary: The closure of a compact set $K$ is also compact.
Proof: An open cover of $\overline K$ is an open cover of $K$, so it has a finite subcover which by the lemma also covers $\overline K$.
