Cantor–Bendixson Theorem Prove that every uncountable closed set $F$ in $R^n$ can be written as $$F=A \cup B$$ where $A$ is a perfect set and set $B$ is a countable set.
$$$$Perfect Set: A set in which all points are accumulation points.
Condensation Point: A point $x$ in $R^n$ is set to be a condensation Point of a set $S$ in $R^n$ if every open n-ball $B(x)$ of $x$ is such that $$B(x) \cap S$$ is uncountable.
$$$$Let $A$ be the set of all condensation points of $F$ lying in $F$ then consider the set $F-A$. If $x$ lies in the set $F-A$ then there exists a open n-ball $B(x)$ of $x$ such that $$B(x) \cap F$$ is countable and hence $$B(x) \cap F-A$$ is countanle. Now for every $x$ in $F-A$ if we chose an open n-ball with the above stated property then this collection of open n-balls is an open covering of $F-A$ and hence by Lindelof's Covering Theorem there exists a countable number of such open n-ball that cover $F-A$ and as each of these open balls contain only a countable number of elements of $F-A$ as stated above we get that the set $F-A$ is countable. So let us chose $$B=F-A$$. Now chose any $x$ in $A$ then for every open n-ball we have $$B(x) \cap F$$ is uncountable. Now suppose there exists a open n-ball $C(x)$ such that $$C(x) \cap A= \phi$$ but as $$C(x) \cap F$$ is uncountable and the set $B=F-A$ is countable so $$C(x) \cap F$$ cannot be uncountable, a contradiction. So every point $x$ in the set $A$ is an accumulation point and the set $A$ is a perfect set.
$$$$Is My Proof Correct??
 A: As others have noted, your proof is at best incomplete: a perfect set is a closed with no isolated points (or, if you prefer, a closed set whose every point is an accumulation point of the set). Your proof that $F\setminus A$ is countable is correct but more complicated than necessary. Your proof that every point of $A$ is an accumulation point of $A$ contains a technical error, and you still have to prove that $A$ is closed.
In proving that $F\setminus A$ is countable there is no need to appeal to the Lindelöf property. $\Bbb R^n$ has a countable base $\mathscr{B}$, so for each $x\in F\setminus A$ we can simply choose a $B_x\in\mathscr{B}$ such that $x\in B_x$ and $B_x\cap F$ is countable. Let $\mathscr{B}_0=\{B_x:x\in F\setminus A\}$; $\mathscr{B}_0\subseteq\mathscr{B}$, so $\mathscr{B}_0$ is countable, and
$$F\setminus A\subseteq F\cap\bigcup\mathscr{B}_0=\bigcup_{B\in\mathscr{B}_0}(F\cap B)\;.$$
That last union is a countable union of countable sets, so it’s countable and therefore so is $F\setminus A$.
To prove that $A$ has no isolated points, you want to show that if $x\in A$, then each open ball centred at $x$ contains a point of $A\setminus\{x\}$, so you let $C(x)$ be an open ball centred at $x$; so far, so good. But then you suppose that $C(x)\cap A=\varnothing$, and this is clearly impossible, since $x\in C(x)\cap A$. If you’re going to argue by contradiction, you should instead assume that $C(x)\cap A=\{x\}$. However, there’s reason to use proof by contradiction here: just observe that if $C(x)$ is an open ball centred at $x\in A$, then $C(x)\cap A=\big(C(x)\cap F\big)\setminus B$, where $C(x)\cap F$ is uncountable and $B$ is countable, so $C(x)\cap A$ is uncountable and therefore certainly contains points of $A$ different from $x$.
To show that $A$ is closed, let $x\in X\setminus A$. $F$ is closed, so if $x\notin F$, then $X\setminus F$ is an open nbhd of $x$ disjoint from $A$. The only other possibility is that $x\in B$, in which case $B_x$ (see above) is an open nbhd of $x$ disjoint from $A$. It follows that $X\setminus A$ is open and hence that $A$ is closed.

It’s possible to modify the proof slightly so that the fact that $A$ is closed comes for free. Simply let $\mathscr{B}$ be a countable base for $\Bbb R^n$, and let $\mathscr{B}_0=\{B\in\mathscr{B}:B\cap F\text{ is countable}\}$. Let $A=F\setminus\bigcup\mathscr{B}_0$; clearly $A$ is a closed set of condensation points of $F$. Moreover, $\mathscr{B}_0$ is countable, so
$$F\setminus A=F\cap\bigcup\mathscr{B}_0=\bigcup_{B\in\mathscr{B}_0}(F\cap B)$$
is countable. The proof that $A$ has no isolated points is the same as before.
