Any infinite set of a compact set $K$ has a limit point in $K$? I'm reading principles of mathematical analysis and have a question about a theorem 2.37.
Theorem 2.37

If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.

The proof is

If no point of $K$ were a limit point of $E$, 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 $E$.
  The same is true of $K$, since $E \subset K$. 
  This contradicts the compactness of $K$.

I understand the first part that states that no finite subcollection of $\{V_q\}$ can cover $E$, in other words, $E$ is not compact. But I don't understand why it means that no finite subcollection can cover $K$. Is the author saying that if a subset of a set K is not compact, then $K$ is not compact? If that's the case, a compact set may have open subsets which are not compact, so I'm confused.
Thanks in advance.
 A: Since for all $q$, $V_q \cap E$ has at most one element, for any finite subset $\{q_1,\ldots,q_n\}$, 
$\bigcup_{i=1}^n V_{q_i} \cap E$ has at most $n$ elements, i.e., is finite.  But $E$ is assumed to be infinite, so we cannot have $\bigcup_{i=1}^n V_{q_i} \supset E$.  Since $K \supset E$, therefore we cannot have $\bigcup_{i=1}^n V_{q_i} \supset K$, contradicting the compactness of $K$.  
A: A set with no limit points is necessarily a closed set.  Being a closed subset of a compact space, it is compact.  On the other hand, you're looking at a proof that $E$ is not compact, so you've got a contradiction.
Alternatively, look at this set of neighborhoods that cover $E$ and add one more open set to this collection: the complement of $E$.  That set is open, since as noted above, $E$ is closed.  Now you've got an open cover of $K$.  It must therefore have a finite subcover.  But every finite subset of this cover fails to cover all of $E$, so again you have a contradiction.
A: Suppose every $ p\in K $ has an open nbhd $ U_p $ such that $ E\cap U_p $ is finite. Then $ C=\{U_p: p\in K\} $ is an open cover of $ K $ but any finite $ D\subset C $ covers only a finite subset of $ E. $ Note that we do not need to assume that $ K $ is a $ T_1 $ space nor even a $ T_0 $ space.
I said "open nbhd" because some people say that a nbhd $ U $ of $ p $ is an open set with $ p\in U, $ while some say that a nbhd $ U $ of $ p $ is any subset of the space, such that $ p\in V\subset U $ for some open $ V.$ 
A: Here is my comment, it maybe a little far from the topic, however, I hope it will be helpful for you.
The compactness condition is too strong, we only need the countable extent condition in the Theorem. 

Countable extent=  The cardinality of any closed discrete subspace must be countable. 

So, in fact, we have 

If $E$ is an infinite subset of a space $K$ which is countable extent, then $E$ has a limit point in $K$.

There are many topological space which is countable extent but not compact. For example, countably compact space, lindelof space etc. It is far from countable extent to compactness.
A: Suppose $\overline{E}$ doesn't have a limit point. Then $\forall e\in E,\exists N_{r_e}(e)\ s.t\ N_{r_e}(e)\cap E=\{e\}$.
Consider the open cover of the compact set ${\overline E}$:
$$\bigcup_eN_{r_e}(e)\rightarrow\bigcup_{i=1}^nN_{r_{e_i}}(e_i)$$
but
$$\bigcup_{i=1}^nN_{r_{e_i}}(e_i)\cap E=\bigcup_{i=1}^n(N_{r_{e_i}}(e_i)\cap E)=\bigcup_{i=1}^ne_i\ne E$$
which is absurd.
