Converging subsequences and subsets having infinite elements 
We have a metric space (V,d). Proof that the following two properties are equivalent.
a) Every sequence $a_n \in V$ has a subsequence which converges to a element $x \in V$
b) For every subset $A\subset V$ with an infinite amount of elements, there exists a $x\in V$ such that for every $\delta>0$ the set $B(x;\delta) \cap A$ has infinite elements.

This exercise has two hints namely:
For a to b: choose a proper sequence $a_n \in A$
For b to a: take a sequence $a_n \in V$ and make a distinction if $\{a_n | n\in \mathbb{N}\}$ has infinite and finite elements
I have absolutely no clue on how to proceed. I find it difficult to already use the hints. So at the moment I'm stuck at finding/choosing a proper sequence.
 A: Expanded HINT: For one direction, assume (a), and let $A$ be an infinite subset of $V$. Since $A$ is infinite, it contains a sequence $\langle a_n:n\in\Bbb Z^+\rangle$ whose points are all distinct. (That is, $a_m\ne a_n$ if $m\ne n$.) By (a) this sequence has a subsequence $\langle a_{n_k}:k\in\Bbb Z^+\rangle$ that converges to some $x\in V$. Use the definition of convergence to show that $A\cap B(x,\delta)$ is infinite for each $\delta>0$; remember, each $a_n$ is in $A$.
For the other direction, assume (b), let $\langle a_n:n\in\Bbb Z^+\rangle$ be a sequence in $V$. There are two cases.


*

*$\{a_n:n\in\Bbb Z^+\}$ is finite; that is, the sequence actually has only finitely many different terms. Show that some term must occur infinitely often, conclude that the sequence must have a constant subsequence, and explain why a constant sequence converges.

*$\{a_n:n\in\Bbb Z^+\}$ is infinite. In this case you can apply (b) to conclude that there are an $x\in V$ such that $B(x,\delta)\cap\{a_n:n\in\Bbb N\}$ is infinite for each $\delta>0$. Use this conclusion to construct recursively a subsequence $\langle a_{n_k}:k\in\Bbb Z^+\rangle$ of $\langle a_n:n\in\Bbb Z^+\rangle$ such that for each $k\in\Bbb Z^+$, $x_{n_k}\in B\left(x,\frac1k\right)$ and $n_k<n_{k+1}$. Then show that $\langle a_n:n\in\Bbb Z^+\rangle$ converges to $x$.
