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 $M$.
Best Answer
$K\setminus O = K \cap (X \setminus O)$ is a closed subset of the compact set $K$ hence compact as well.