claim: Let $(M, d)$ be a metric space and $K \subset M$ compact, $O \subset M$ open. Show that $K - O$ is compact.

Proof: I think this should follow directly. If $K$ is a compact set, that means every open cover of $K$ contains a finite subcover.

let's define an arbitrary finite subcover of $K$ to be $\{A\} = A\_1 \cup A\_2 \cup \cdots \cup A\_n$

Since $K - O$ is simply the set $K$ with the elements of $O$ taken out, that means our set we need to cover is smaller, and therefore surely covered by $\{A\}$.

Since $K - O$ is covered by $\{A\}$ and $\{A\}$ is a finite subcovering by construction, then $K - O$ is compact.

Does this work?

I didn't even use the fact that $O$ was an open set in $M$.