Let $F \subset \mathbb{R}^n$ be a nonempty countable or finite closed set. Prove that $F$ must have an isolated point. I tried to think of the problem using contradiction that if $F$ does not have any isolated point (that is, $F$ contains only limit points) then $F$ is uncountable. But then I was stuck into how to prove that $F$ is uncountable. Can anyone provide a method to proving this proposition?
 A: Assume that $F$ is a closed set without any isolated point. 
Choose any $x_0 \in F$, since it's not an isolated point we can find a ball $B_0$ centered at $x_0$ so that $B_0$ contains at least $2$ distinct points other than $x_0$ itself, let's call them $x_{00}$ and $x_{01}$. Let $B_{00}$ and $B_{01}$ be disjoint ball centered at $x_{00}$ and $x_{01}$ respectively, each having radius less than half of $B_0$.
We do the same for $x_{00}$ and $x_{01}$ to find distinct points $x_{000}, x_{001}, x_{010}$ and $x_{011}$ with their respective disjoint balls. More generally, for any (finite) binary index $\alpha=a_1a_2\dots a_n$, we can find $x_{\alpha 0}$ and $x_{\alpha 1}$ in $B_\alpha$, and two disjoint balls $B_{\alpha 0}$ and $B_{\alpha 1}$ centered at these points, each having radius less than half of the radius of $B_\alpha$.
Each sequence $x_{a_1}, x_{a_1 a_2}, x_{a_1 a_2 a_3}, \dots$ is convergent (since the radius of $B_{a_1}, B_{a_1 a_2}, B_{a_1 a_2 a_3}, \dots$ goes to $0$), and hence determines a unique point in $F$ because $F$ is closed. Uniqueness comes from the fact that we chose the balls $B_{\alpha 0}$ and $B_{\alpha 1}$ to be disjoint.
There are as many such sequences as the number of all binary digits (beginning with $0$), i.e. as many as the number of elements in the set
$$
S=\{ (a_1, a_2, a_3, \dots) : a_i\in\{0,1\}\  \}.
$$
It is well known that $S$ is uncountable since there's a bijection between $S$ and the unit interval $[0,1]$. The idea is to write each element in $[0, 1]$ as a binary digits.
A: Endow $F$ with the induced metric. It is complete. Suppose there does not exist an isolated point. Write $F=\{x_1,...,x_n,....\}$. Let $U_n$ be the complementary subspace of $\{x_1,....,x_n\} $. It is dense in $F$ since there does exist isolated point. Baire theorem implies the intersection $\cap U_n$ is dense in $F$. Contradiction. Since it is empty. 
