I read this in Theorem 2.35 of Baby Rudin:
Corollary. In the context of metric spaces) If $F$ is closed and $K$ is compact then $F \cap K$ is compact.
Proof. Because intersections of closed sets are closed and because compact subsets of metric spaces are closed, so is $F \cap K$; since $F \cap K \subset K$, theorem 2.35 shows $F \cap K$ is compact.
He assumes that $F \cap K$ is a compact subset in order to prove $F \cap K$ is compact.