Rudin's Exercise 2.27 in Principles of Mathematical Analysis: Why are countable many points of $E$ not in perfect set $P$? Here is the theorem to be proved:
Suppose $E \subset \mathbb{R}^k$, $E$ is uncountable, and $P$ is the set of condensation points of $E$.  Prove that $P$ is perfect and that there are at most countably many points in $E$ that are not in $P$. 
I tried to prove it as follows:  
Let $p\in \mathbb{R}^k$ be a limit point of $P$, then for some $\delta \gt 0, N_\delta(p)$ contains $y\ne x$ such that $y\in P$. Let $r=\frac{1}{2} \min (d(p,y), \delta-d(p,y))$. We have $N_r(y)\subset N_\delta (p)\implies N_r(y) $ contains uncountable many points of $E$ (as $y$ is condensation point)$\implies N_\delta (p)$ contains uncountable many points of  $E\implies p$ is a condensation point of $E$. So $p\in P$. Hence $P$ is closed. $\tag{1}$
Let $p\in P$ be an isolated point of $P$. $\exists\delta \gt 0: N_\delta (p)\cap P=\{p\}$.  Since $p\in P$, we have that $N_\delta (p)\cap E$ is uncountable. 
Suppose that $N_\delta (p)$ does not contain any other condensation point of $E$ except $p$. 
Then for every $y\in N_\delta (p)-\{p\},\exists r\gt 0: N_r (y)$ is at most countable for every $y\in N_\delta (p)-\{p\}$. 
We have $N_\delta (p)\subset (\cup_{y\in N_\delta (p)-\{p\}} N_r(y))\cup\{p\}$, whence it follows that $ E\cap N_\delta (p)\subset (E\cap \{p\})\cap_{y\in N_\delta (p)-\{p\}} (E\cap N_r(y)\subset E\cap N_r(y)$ for some $y\in N_\delta (p)-\{p\}$. Hence, $ N_\delta(p)$ is at most countable, which is a contradiction. Therefore, all points of $P$ are limit points of $P$. $\tag{2}$ 
By $(1)$ and $(2), P$ is perfect. 
I got stuck at proving the second part of this question. 
I tried to use hint provided in the exercise: "Let $\{V_n\}$ be a countable base of $\mathbb R^k$, let $W$ be the union of those $V_n$ for which $E\cap V_n$ is at most countable and show that $P=W^c$."
I understand that since $\mathbb R^k$ is separable, it must have a countable base. I don't know how to proceed further using the hint. I also don't understand why such $W$ exists. Can existence of such $W$ be proven? 
Please help. Thanks.
 A: In your proof that $P$ is closed you need to say that for every $\delta>0$ the nbhd $N_\delta(p)$ contains a point of $P\setminus\{p\}$, not just some $\delta>0$; otherwise you cannot conclude that $p\in P$. Otherwise that proof is correct, though you could also simply let $r=\delta-d(p,y)$: that still ensures that $N_r(y)\subseteq N_\delta(p)$. However, it is simpler to prove that $\Bbb R^k\setminus P$ is open. If $x\in\Bbb R^k\setminus P$, there is an $\epsilon>0$ such that $N_\epsilon(x)\cap E$ is countable. $N_\epsilon(x)$ is an open nbhd of each of its points, so every point of $N_\epsilon(x)$ has an open nbhd that contains only countably many points of $E$, and therefore no point of $N_\epsilon(x)$ is a condensation point of $E$; in other words, $N_\epsilon(x)\cap P=\varnothing$. Since $x\in\Bbb R^k\setminus P$ was arbitrary, it follows that $\Bbb R^k\setminus P$ is open and hence that $P$ is closed.
Your argument to show that $P$ has no isolated points, however, is not correct: somehow you turned
$$E\cap\bigcup_{y\in N_\delta(p)\setminus\{p\}}N_r(y)$$
into
$$\bigcap_{y\in N_\delta(p)\setminus\{p\}}\big(E\cap N_r(y)\big)\,,$$
and the two sets are definitely not equal in general. I should also note that you cannot use the same $r$ for every $y\in N_\delta(p)\setminus\{p\}$: you need to index the $r$s by $y$, e.g., as $r(y)$ or $r_y$, since each point $y$ may require a different $r$.
The easiest way to prove that $P$ has no isolated points is to use the same idea that is used to prove that $E\setminus P$ is countable. (Note: I use countable in its more usual sense of finite or countably infinite.) Let $\mathscr{B}=\{B_n:n\in\Bbb N\}$ be a countable base for $\Bbb R^k$. Suppose that $p$ is an isolated point of $P$; then there is an $m\in\Bbb N$ such that $B_m\cap P=\{p\}$. Thus, for each $x\in B_m\setminus\{p\}$ there is a $B_{n(x)}\in\mathscr{B}$ such that $x\in B_{n(x)}\subseteq B_m$, and $B_{n(x)}\cap E$ is countable. Let $N=\big\{n(x):x\in B_m\setminus\{p\}\big\}$; then $N$ is countable, since it’s a subset of $\Bbb N$, and
$$E\cap\big(B_m\setminus\{p\}\big)=\bigcup_{x\in B_m\setminus\{p\}}(E\cap B_{n(x)})=\bigcup_{n\in N}(E\cap B_n)\,.$$
That last union is the union of countably many countable sets, so it is countable. Thus, $B_m$ is an open nbhd of $p$ that contains only countably many points of $E$, contradicting the assumption that $p\in P$. It follows that $P$ has no isolated points.
Exactly the same idea is used to show that $E\setminus P$ is countable. Let
$$N=\{n\in\Bbb N:E\cap B_n\text{ is countable}\}\,,$$
and let $W=\bigcup\limits_{n\in N}B_n$. Clearly
$$E\cap W=E\cap\bigcup_{n\in N}B_n=\bigcup_{n\in N}(E\cap B_n)$$
is the union of countably many countable sets, so $E\cap W$ is countable. It’s also clear that $W\cap P=\varnothing$: if $x\in W$, $W$ is an open nbhd of $x$ containing only countably many points of $E$, so $x$ is not a condensation point of $E$.
Finally, suppose that $x\in\Bbb R^k\setminus W$, and let $U$ be any open nbhd of $x$. $\mathscr{B}$ is a base for $\Bbb R^k$, so there is an $m\in\Bbb N$ such that $x\in B_m\subseteq U$. For each $n\in N$ we have $B_n\subseteq W$, and $x\in B_m\setminus W$, so $m\notin N$. By the definition of $N$ this means that $E\cap B_m$ is uncountable, and therefore $E\cap U$ is uncountable (since $B_m\subseteq U$). Thus, $x\in P$, and therefore $\Bbb R^k\setminus W\subseteq P$, i.e., $W\cup P=\Bbb R^k$. We already saw that $W$ and $P$ are disjoint, so they must be complementary: $P=\Bbb R^k\setminus W$, and therefore $E\setminus P=E\cap W$, which we’ve already shown to be countable.
Note, by the way, that this argument also shows directly that $P$ is closed, since it’s the complement of an open set.
