The below is the proof for the theorem that if $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.
The only part I can't accept in the proof is the last sentence.
So, it's shown that no finite subcollection of $\{V_q\}$ can cover $K$.
However, how does this contradict the compactness of $K$?
Thank you!
$\mathbf{2.37}\,\,$ *Theorem*$\,\,\,\,$*If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.*
*Proof*$\quad$If no point of $K$ where a limit point of $E$, then each $q\in K$ would have a neighborhood $V_q$ which contains at most one point of $E$ (namely, $q$, if $q\in E$). It is clear that no finite subcollection of $\{V_q\}$ can cover $E$; and the same is true of $K$, since $E\subset K$. This contradicts the compactness of $K$.
Best Answer
Each $q \in K$ has a neighbourhood $V_q$ that contains only finitely many points of $E$. The family $\left(\overset{\circ}{V}_q\right)_{q\in K}$ is an open cover of $K$, hence has a finite subcover, since $K$ is compact.
That contradicts the fact that $E$ is infinite, since any finite subfamily of the $V_q$ can only cover finitely many points of $E$.