Closure of a compact set in Regular space $ X $

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 $C$.

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 $\overline K$.