Limit point of an infinite subset of a compact set I have a doubt about the proof of theorem 2.37 in Rudin's Principle of Mathematical Analysis, which I have included below.
My problem is, the proof treats a collection of singletons $\{V_q\}$ as an open cover; however, finite sets are closed (edit: "and also not open" is what I meant to say; thanks to GNU Emacs for pointing this out) which means the collection cannot be an open cover. If the collection is not an open cover, then the contradiction falls apart. So, what is wrong with my reasoning here?

2.37 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 $E$ (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 $K$.

 A: $V_q$ is just some open set containing $q$ such that $V_q\cap E=\{q\}$; $V_q$ may (probably will) have other elements. The issue I suspect is the phrase

which contains at most one point of $E$.

This is not saying that $V_q$ contains at most one point! Just that $V_q$ contains at most one point in $E$ - $V_q$ is allowed to contain lots of points not in $E$.
A: The sets $V_q$ are not necessarily singletons: we only know that $V_q\cap E$ is either a singleton or empty, but $V_q$ can contain many other points that aren't in $E$.
But even if they were singletons, the proof would still work!  The proof is a proof by contradiction: assuming that $E$ has no limit point in $K$, you reach a contradiction.  The proof produces a set $V_q$ which is by definition open: namely, it is an open set containing $q$ which contains at most one element of $E$ (such an open set exists since $q$ is not a limit point of $E$).  If you could prove that $V_q$ is finite and thus could not actually be open, this is a contradiction.  But that's fine, since we were trying to get a contradiction anyways!
A: Thank you all for your help. I've just realised that I forgot a very simple fact that every neighbourhood is an open set (theorem 2.19). Thus, even if $V_q$ contains only one element $q$, then ${q}$ is rather trivially an interior point of $V_q$. 
Conclusion: Since theorem 2.19 implies $V_q$ is an open set and the union of $\{V_q\}$ contains $K\supset E$ (as defined in theorem 2.37), then $\{V_q\}$ is an open cover of $K$. However, it is clear that no finite subcollection of $\{V_q\}$ can cover $E\subset K$; and, thus, this contradicts compactness of $K$, and the theorem holds.
Remarks: 
My statement above the conclusion is the converse of the definition of an open set. Yet, I still decided to make the statement due to the proof of theorem 2.19: it shows that for a neighbourhood $N$ of $p$ and any point $q\in N$, then $q$ is an interior point. Thus, because the aforementioned set $V_q$ is open, then it has, as its only element, an interior point.
It's a rather subtle remark, but I think it's a necessary one to make for those who are thinking that my statement above the conclusion is the converse of the definition of an open set, which in turn needs a further explanation.
I'd like to add further that the only problem I had was wether or not each $V_q$ was open. However, it seems that the choice of $\{V_q\}$ has gotten more attention. I do not have a problem with this because the choice implies either $V_q$ contains only one $q\in E$, or $q$ is an interior point of $K\setminus{E}$ such that $V_q\cap{E}=\emptyset$. Therefore, no finite subcollection of $\{V_q\}$ can cover $E$.
