Every closed set in a separable metric space is the union of a perfect set and a set which is at most countable 
Prove that every closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable. 

(Rudin's Principles of Mathematical Analysis, 3rd ed., Chapter 2, Problem 28)
Now I've managed to hit upon a proof using the preceding problem (which can be generalised) as follows: 
Let $X$ be a separable metric space, $E \subset X$, $E$ is uncountable, and let $P$ be the set of all condensation points of $E$. Prove that $P$ is perfect and that at most countably many points of $E$ are not in $P$. In other words, show that $P^c \cap E$ is at most countable. 
Now what bothers me is what Rudin states as the corollary to the result he asks us to prove in Problem 28. Here's what he states: 
Every countable closed set in $\mathbb{R}^k$ has isolated points. 
How to prove this from the original result? 
Can we also state the following? 
Every countable closed set in a separable metric space has isolated points. 


Every countable closed set in $\mathbb{R}^k$ has isolated points. 

HINT: Negate it (that is there is no isolated point) then surround two different points by a closed ball then pick $2$ different points inside each ball surround again the $4$ by a smaller closed ball (that is none of the points is inside any of the others balls), and repeat, then apply the Nested Intersection Theorem (Remember a closed subset of a complete space is complete) to any sequence of nested balls and deduce you have a bijection of your limit points with $\{0,1\}^{\aleph} \sim\aleph^{\aleph}$
EDITED:
Suppose your set doesnt have any isolated points, then it is a perfect set of $\mathbb{R}^k$ but this would mean that is not countable, contradiction.
