Let $I$ be an indexing set and $F=\{V_{\alpha}\mid\alpha\in I\}$ be an arbitrary open cover for $C$. Since $X\setminus C$ is open, it follows that $F$ together with $X\setminus C$ is an open cover for $K$. Thus, $K$ can be covered by a finite number of sets, say, $V_{1},\ldots,V_{N}$ from $F$ together with possibly $X\setminus C$. Since $C\subset K$, $V_{1},\ldots,V_{N}$ cover $C$, and it follows that $C$ is compact. ∎

Let $I$ be an indexing set and $\{A_{{\alpha}}\}_{{\alpha\in I}}$ be a collection of $X$-closed sets contained in $C$ such that, for any finite $J\subseteq I$, $\displaystyle\bigcap_{{\alpha\in J}}A_{{\alpha}}$ is not empty. Recall that, for every $\alpha\in I$, $A_{{\alpha}}\subseteq C\subseteq K$. Thus, for every $\alpha\in I$, $A_{{\alpha}}=K\cap A_{{\alpha}}$. Therefore, $\{A_{{\alpha}}\}_{{\alpha\in I}}$ are $K$-closed subsets of $K$ (see this page) such that, for any finite $J\subseteq I$, $\displaystyle\bigcap_{{\alpha\in J}}A_{{\alpha}}$ is not empty. As $K$ is compact, $\displaystyle\bigcap_{{\alpha\in I}}A_{{\alpha}}$ is not empty (again, by this result). This proves the claim. ∎