Show that a metric space $V$ is sequentially compact iff it is limit point compact Given is a metric space (V,d). Show that the following properties are equivalent:
a) V is sequential compact ( so each sequence in V has a convergent subsequence witha limit in V)
b) for each subset $A \subset V$ with an infinite amount of elements, there exists a point $ x \in V$ such that for each $ \delta >0$ , $B(x; \delta) \cap  A$ 
contains an infinite amount of elements.
My idea for $a \Rightarrow b$ was to pick a suitable sequence $a_n$ in A, and we know $a_n$ converges, because V is sequential compact, than show that there exists 
a point $ x \in V$ such that for each $ \delta >0$ , $B(x; \delta) \cap  A$ 
contains an infinite amount of elements. 
For $b \Rightarrow  a$ I wanted to take a sequence $a_n$ out of $V$, and look at two cases, one with {$a_n|n\in N$} with a finite amount of elements, and one with an infinite amount. And from there show that V is sequential compact.
I don't really know if these ideas are right, and if so how to continue. 
I also have some difficulty with picking suitable sequences, like in $a \Rightarrow b$ , tips about that are much appreciated aswel!

choose $a_n \in B(x;\frac{1}{n})∩A$
