Studying on my own currently and having trouble understanding a topology proof.

Theorem: Compact subsets of metric spaces are closed

Proof

Let $\displaystyle K$ be a compact subset of a metric space $\displaystyle X$. We shall prove that the complement of $\displaystyle K$ is an open subset of $\displaystyle X$

Suppose $\displaystyle p \in{X}$, $\displaystyle p$ $\displaystyle \notin$ $\displaystyle K$. If $\displaystyle q\in{K}$, let $\displaystyle V_q$ and $\displaystyle W_q$ be neighbourhoods of $\displaystyle p$ and $\displaystyle q$ respectively, of radius less than $\displaystyle \frac{1}{2}d(p,q)$

Since $\displaystyle K$ is compact, there are finitely many points $\displaystyle q_1,...,q_n$ in $\displaystyle K$ such that

$\displaystyle K\subset{W_{q_1}\cup...\cup{W_{q_n}}=W$

If $\displaystyle V=V_{q_1}\cap...\cap{V_{q_n}}$, then $\displaystyle V$ is a neighbourhood of $\displaystyle p$ which does not intersect $\displaystyle W$. Hence $\displaystyle V\subset{K^c}$, so that $\displaystyle p$ is an interior point of $\displaystyle K^c$. The theorem follows.

Question

I am confused as to how $\displaystyle V_q$ and $\displaystyle W_q$ are defined

Is it that

$\displaystyle W_{q_i}:[\forall{p}: d(p,q_i)<\frac{d(p,q)}{2}]$

$\displaystyle V_{q_i}:[\forall{q}: d(p,q_i)<\frac{d(p,q)}{2}]$

Also, how does the chosen radius play a role in the proof?