# Proof of Rudin theorem 2.34 compact subsets of metric spaces are closed

Rudin writes at a certain point in the proof:

$\textbf {Since$K$is compact}$, there are finitely many points $q_1,...,q_n$ in $K$ such that:

$K \subset W_{q1} \cup W_{q2} \cup ... \cup W_{qn}$ where the $\textbf{$W_{qi}$'s are neighborhoods}$ of a point $q\in K$

Well the implication in bold is my problem, that's definitely not what the definition of a compact set states.

The definition says that for every open cover we can get an open finite subcover but we can't assume that they're neighborhoods right?? How does he make this jump??

Thanks for the help!

The definition of a compact set means that every open cover of $K$ has a finite cover. For each $q \in K$, choose a neighborhood $W_q$ of $K$; then $\{W_q : q \in K\}$ is an open cover of $K$, since it clearly contains every point in $K$.
Hence, we may select a finite subcover, and just let $q_1, ..., q_n$ be the points corresponding to the open sets in our finite cover.
The open cover is probably defined semi-explicitely just before the part that you have cited and consists of the sets $W_q$ for all $q\in K$.