Baby Rudin exercise 2.27: why is $P$ perfect? The original problem goes as follows (reloaded the image)

I have solved the large part of this problem (already shown that $P=W^c$ and that $P$ is closed), but I just cannot figure out a proof showing that every neighbourhood of a condensation point must contain another condensation point.
What I tried: suppose $p\in P$ is an isolated point of $P$ so there would be an neighbourhood of $p$, say, $N_r(p)$ that contains no other point of $P$ than $p$ itself. In other words, $N_r(p)$ only contains points that are either isolated points of $E$ or its $\aleph_0-$ accumulation points. But I could not make any contradiction out of this hypothesis. 
One answer I have at hand simply claims effortlessly that: 

if $p$ is an isolated point of $P$ then there would be a neighbourhood $N$ of $p$ such that $N\cap E=\varnothing$. 

I think this explanation is, if not sheer nonsense (by which I mean: the author might have treated "isolated in $P$" wrongly as "isolated in $E$" so he/she was able to draw such a non-trivial conclusion so effortlessly?), at least over-simplified for me to understand. 
Could you enlighten  me? Thanks!
 A: Suppose that $p$ is an isolated condensation point of $E$, so there is an open set $B$ containing $p$ such that $B\cap P=\{p\}$. Every point $q\ne p$ in $B$ has an open neighborhood $A_q\subseteq B$ such that $A_q\cap E$ is at most countable. The open covering $\mathcal A=\{A_q\}$ of $B\setminus\{p\}$ admits an at most countable subcover, say $\mathcal D\subseteq\mathcal A$; but then $\bigl(B\setminus\{p\}\bigr)\cap E=\cup_{D\in\mathcal D}(D\cap E)$ is at most countable, which is absurd.
A: Note : I define "countable" as "not uncountable" so countable means "finite or countably infinite".....We may define $W$ as the union of all open sets in $R^n$ that have countable intersection with $E$..... Let  $S=\{n \in N : E\cap V_n \text{ is countable }\}$. The crux is that  $E\cap W$ is countable because $E \cap  W=\cup_{n \in S}(E\cap V_n)$ is a countable union of countable sets; and that any open subset of $R^n$ that has countable intersection with $E$ is a subset of $W$; And that $ E\backslash W \subset P$....... Now $ P=R^n\backslash W$.... So if $p\in P$ and if $U$ is a neighborhood of $p$ in $R^n$ then $U\cap E$ is uncountable (otherwise $p\in U\subset W$,contradicting $p \not \in W$)....... $$\text{But the uncountable set }  U\cap E =$$ $$(U\cap (E\cap W)\cup (U\cap (E\backslash W)=$$ $$ (U\cap (E\cap W)\cup (U\cap (E\cap  P))$$ and $(U\cap (E\cap W)$ is only countable because it is a subset of the countable set $E\cap W$. Therefore $(U\cap P)$ must be  uncountable. 
