A 2nd countable metric space is separable 
A 2nd countable metric space is separable

Can anyone give an idea on how to prove that, I could prove the converse quite easily but I am not sure which set to consider in order to show that it is dense.
 A: HINT: Start with a countable base $\mathscr{B}$ for the topology and pick one point from each member of $\mathscr{B}$.
A: I am sorry. As you only asked for a hint, I now hide my answer in the following spoiler box:

 Let $(U_n)_{n \in \mathbb{N}}$ be a countable basis of nonempty sets for the 2nd countable metrik space $(X,d)$. Fix a choice function $f: \bigcup_{n \in \mathbb N} U_n \mapsto V$, where $V$ is the universal class. We claim that the set $(f(U_n))_{n \in \mathbb N}$ is dense. In fact, for any open $U \subseteq X$ there is an $n_0 \in \mathbb{N}$ such that $U_{n_0} \subseteq U$ (by definition of a basis) and thus $f(U_{n_0}) \in U$, showing density.

A: You need to use the axiom of choice, you can't always write an explicit countable dense subset.
Hint: Use the fact that you have a countable basis to the topology, and those sets are all non-empty.
A: The metric assumption is really useless, it works in every second countable topological space: pick an element from each open basic set (using AC) and collect them in a set. This countable set is dense in the original topological space.
