Begging the question in Rudin? 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.

I have a Second Edition (1964) of Rudin in which the proof is given this way:

Theorems $2.26(b)$ and $2.34$ show that $F\cap K$ is closed; since
  $F\cap K \subset K$, Theorem $2.35$ shows that $F\cap K$ is compact.

Theorem $2.26(b)$ says that intersections of closed sets are closed, $2.34$ says that compact subsets of metric spaces are closed, and $2.35$ that closed subsets of compact sets are compact.
