If $B$ is the set of condensation points of a set $A$, show that every point of $B$ is a condensation point of $B$. This problem comes from Dieudonné's Foundations of Modern Analysis:

Let $E$ be a separable metric space. A condensation point $x$ of a subset $A$ of $E$ is a point $x \in E$ such that in every neighborhood of $x$, there is a nondenumerable set of points of $A$. Show that: (a) If $A$ has no condensation point, it is denumerable (consider the intersections of $A$ with the sets of a basis for the open sets of $E$).
(b) If $B$ is the set of condensation points of a set $A$, show that every point of $B$ is a condensation point of $B$, and that $A \cap (\complement B)$ is at most denumerable. (Observe that $B$ is closed, and use (a).)

I've managed to solve (a) in the following way:
Let $x$ be any point of $A$. There exists a neighbourhood $V_x$ of $x$, such that it contains at most $\aleph_0$ elements from $A$ (since $A$ has no condensation points). Since $E$ is separable, there exists a denumerable basis $(G_\lambda)$ for all open sets of $E$, and there exists an element of the basis $G_x$ such that $x \in G_x \subset V$. Now we have $A \subset \bigcup_{x \in A} G_x$. But since every set $G_x$ contains at most $\aleph_0$ elements of $A$ (as it is a subset of $V_x$) and the sum $\bigcup_{x \in A} G_x$ is a sum of at most $\aleph_0$ distinct sets, $\bigcup_{x \in A} G_x$ contains itself at most $\aleph_0$ elements of $A$, and thus $|A| \leq \aleph_0$.
However, I can't tackle (b), even with the hint. I don't really feel how closeness of $B$ is supposed to help us here.
 A: For part (a): Your first part may not be correct. Since you don't apriori know about countability (or uncountability of $A$), you don't know apriori whether  $ \bigcup_{x \in A} G_x$ is a countable union or uncountable union. (Note that countable union of countable sets is necessarily countable).
Therefore, you may proceed as follows for this part: 
Let $(G_n)$ be the countable base of $E$. 
Since $A$ does not have any condensation points, if $A$ is empty, we are done. If $A$ is non empty, then let $N=\{k\in \mathbb N: G_{n_k}\cap A \text{ is countable }\}$.
Define $S=\cup_{k\in N} G_{n_k}$.  Let $E\setminus S\ne\emptyset $. Then for any $x$ in $E\setminus S$, if $B(x,\delta)$ is any arbitrary open ball around it then there exists a $G_{n_{k'}}\in (G_n)$ such that $x\in G_{n_{k'}}\subset B(x,\delta)$ and clearly
$n_{k'}\notin N$. It follows that $x$ is a condensation point of $A$, which is a contradiction. Hence $E\setminus S=\emptyset$. It follows that $E=S\implies A$ is countable. 
For part $(b)$: First of all, note that $B$ is closed. If not, let $cl(B)-B\ne \emptyset$. 
Let $x\in cl (B)-B$. It follows that $x$ is not a condensation point of $A$ and hence $\exists \delta\gt 0$ such that $B(x,\delta)\cap A$ is countable. 
Since $x$ is a limit point of $B, B(x,\delta)\cap B$ is infinite. Choose $y\in B(x,\delta)\cap B$ such that $y\ne x$ and take $r=\frac 12\min\{d(x,y),\delta -d(x,y)\}$ so that $B(y,r)\subset B(x,\delta) \implies B(y,r)$ contains countably many points of $A$ which is a contradiction as $y$ is a condensation point of $A$.  Therefore, $cl (B)=B\iff B$ is closed. It follows that $B^c$ is open.
Now let $C= \text{set of condensation points of }A\cap B^c$.  
Claim: $C$ is empty. 
Proof: If not, suppose that $C\ne \emptyset$. Let $p\in C$ be any point in $C$.  For any $\epsilon \gt 0, B(p,\epsilon)\cap (A\cap B^c)\subset B(x,\epsilon)\cap A\implies B(p,\epsilon)\cap (A\cap B^c)$ is countable which is a contradiction as $p$ is a condensation point of $A\cap B^c$. Hence it follows by contradiction that $C=\emptyset\implies A\cap B^c$ is countable.
