Suppose E is an infinite subset of a metric space X. Prove that x is a limit point of $E$ if and only if there is a sequence $\left \{ x_n \right \}^\infty_{n=1} \subset E $ that converges to x. 
This was part of our practice final and I have no idea how to solve it. I feel like we have to use covers and subcovers, but can anyone help?    
 A: A correct statement is that $x$ is a limit point of $E$ iff there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $E\setminus\{x\}$ that converges to $x$. You should have little difficulty showing that if such a sequence exists, then $x$ is a limit point of $E$; just use the definitions of limit point and limit of a sequence.
The slightly harder direction is to show that if $x$ is a limit point of $E$, then there is such a sequence. Let $n\in\Bbb N$; since $x$ is a limit point of $E$, there is a point $x_n\in B(x,2^{-n})\cap(E\setminus\{x\})$, i.e., a point of $E$ such that $0<d(x_n,x)<2^{-n}$. Now just show that $\langle x_n:n\in\Bbb N\rangle$ converges to $x$.
Note that instead of $2^{-n}$ I could have used any sequence of positive real numbers converging to $0$. That is, if $\langle r_n:n\in\Bbb N\rangle$ is a sequence of positive real numbers, then for each $n\in\Bbb N$ there must be a point $x_n\in B(x,r_n)\cap(E\setminus\{x\})$, and it’s straightforward to show that $\langle x_n:n\in\Bbb N\rangle$ converges to $x$, simply using the fact that for any $\epsilon>0$ there is an $n\in\Bbb N$ such that $r_n<\epsilon$.
