Prove that a subset of a separable set is itself separable The problem statement, all variables and given/known data:
Show that if $X$ is a subset of $M$ and $(M,d)$ is separable, then $(X,d)$ is separable. [This may be a little bit trickier than it looks - $E$ may be a countable dense subset of $M$ with $X\cap E = \varnothing$.] 
Definitions
Per our book:
A metric space $(M,d)$ is separable if there exists a countable dense $E$ contained in $M$. 
$E$ contained in $M$ is dense if $\forall m\in M$, $\forall ε>0\in\mathbb R$, $\exists e\in E$ s.t. $d(m,e) < ε$ 
The attempt at a solution
My best attempt was doomed from the start, because I don't quite understand the hint. My thought process went as follows:
Since $X$ is a subset of $M$, $\forall x\in X, x\in M$. Thus, since $E$ is dense in $M$,
$\forall x\in X$, and ε > 0, $\exists e\in E$ st $d(x,e)<ε $. 
At this point, I was done, because the set of $e$'s satisfying the above, is a subset of $E$, a countable set. So a subset of a countable set is dense in $X$, and $X$ is separable. This is incorrect, but I cannot see why. 
Any help clearing up the confusion would be greatly appreciated.
Thanks!
Edit: I wish I could upvote all of you for your help! I really appreciate the speedy replies and attempts to make this information clear to me. 

Hint (for a different proof strategy): Show that $(M,d)$ is second countable, i.e. there is a countable collection $\mathcal B = \{U_n : n \in \omega\}$ of open sets of $M$ such that any open set $U$ in $M$ can be written as a union of elements of $\mathcal B$. Then show that that $X$ also is second countable. And finally show that second countable implies separable.
