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.
 A: That's not what Rudin says. He says that since $\;F\cap K\;$ is closed [as an intersection of closed sets] and $\;F\cap K\subset K\;$ and $\;K\;$ is compact, then so is $\;F\cap K\;$ .
A: Claim: Compact subsets of metric spaces are closed.
Proof. Suppose the compact set $K$ is not closed. Then there is a limit point $x \notin K$ such that $x_n \in K$ and $x_n \to x$ with respect to the metric.
Consider the open cover which are rings around $x$. That is, consider a sequence $\{r_n\}_{n=-\infty}^\infty$ with $r_n < r_{n+1}$. Let as well $r_n \to 0$ as $n \to -\infty$ and $r_n \to \infty$ as $n \to \infty$. Take $O_n = B_{r_{n+1}}(x)\setminus B_{r_{n-1}}(x)$. Let the open cover be then $K \subseteq \cup_{n=-\infty}^\infty O_n$.
Since $K$ is compact, there is a finite subcover. But then we exclude $O_{m}(x)$ for some smallest $m$, and hence $B_{r_{m-1}}(x) \cap K = \emptyset$. But this contradicts that $x$ is a limit point.
A: 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.
