Rudin's theorem on compact sets and limit points I have trouble in understanding the proof of a theorem on compact sets and limit points in Rudin's book.

Theorem: If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.
Proof: If no point of $K$ were a limit point of $E$, then each $q\in K$ would have a neighborhood $V_q$ which contains at most one point of $E$. It is clear that no finite subcollection of $\{V_q\}$ can cover $E$, and thus $K$. This contradicts the compactness of $K$.

My question is: How do we know the collection $\{V_q\}$ is an open cover of $K$, given $V_q\cap E=\emptyset$ or $q\in E$? I mean, the radius of each open neighborhood around $q$ is fixed, because each punctured neighborhood has to satisfy certain conditions $V_q^* \cap E=\emptyset$, where $*$ denotes "punctured." In this case, how can we ensure the collection of such neighborhoods still covers $K$?
 A: Since $V_q$ is a neighborhood of $q$, $q\in V_q$. And so$$K=\bigcup_{q\in K}\{q\}\subset\bigcup_{q\in K}V_q.$$
A: Correct me  if wrong.
A bit of context:
Let $X$ be a metric space.
$E \subset K \subset X$, $K$ compact, $E$ infinite.
Definition:
$p \in K$  is a limit point of $E$ if every neighbourhood of $p$ contains a point  $q \not = p$ where $q \in E$.
Negation: $p$ is not a limit point of $E$:
There is a neighbourhood of $p$ that does not contain a $q \not = p$,  $q \in E$.
Either:
1) $p \not \in E$ , then there is a $V_p$ such that 
$V_p \cap E =\emptyset.$
Or 
2) $p \in E$, then $V_p \cap E = {p}$.
Recall $E \subset K =\bigcup_{p} V_p$, $p \in K$.
Since $E$ is infinite, no finite subcollection can cover $E$, contradiction to $K$ compact.
P.S. Short proof of the above statement.
Since $K$ compact there is a finite subcollection 
$E \subset K \subset \bigcup_{i} V_{p_i}$ , $i=1,2,...n.$
Recall;  For every $i=1,2,...,n$, $V_{p_i}$ has at most $1$ element of $E$.
Hence a contradiction to $E$ is infinite.
