If $X$ is a separable metric space and $M \subset X$ is a metric subspace then $M$ is separable. If $X$ is a separable metric space and $M \subset X$ is a metric subspace then $M$ is separable.
So i'm having the following doubt. Since $X$ is separable, there exist $\mathcal{D} \subset X$ dense and countable. So I want to show that $M\cap \mathcal{D}$ is dense in $M$ (and of course countable). So, when taking $x \in M$ and $\epsilon>0$ my following question arises. I should see that $B(x,\epsilon) \cap (M\cap \mathcal{D}) \neq \emptyset$ or $(B(x,\epsilon)\cap M) \cap (M\cap \mathcal{D}) \neq \emptyset$, more generally, the density of $M \cap D$ implies not null intersection with every open set in $M$? 
 A: Just so there is the option to remove this question from the unanswered queue, here is a post consolidating the info in the comments above.
A key thing to grasp is that, for general topological spaces, separability is not a hereditary property (a topological property is called hereditary if, whenever a topological space $X$ has the property, so does every subspace $Y \subseteq X$). Here is an example showing that separability is not necessarily inherited by a subspace.

Example: Let $I$ denote the closed unit interval and let $X = I^I$, the product of continuum-many copies of $I$. The space $X$ is separable and, by Tychonoff's theorem, a compact Hausdorff space to boot. An example of a countable dense subset $D \subseteq X$  is the set of all rational valued functions $I \to I$ which are piecewise-constant with respect to a rational partition of $I$. Another example of a countable dense subspace is the set of all maps $I \to I$ which are described by a polynomial with rational coefficients.  An example of a nonseparable subspace $Y \subseteq X$ is the set of all characteristic functions of points. That is, the set of all functions $\chi_p : I \to I$, $p \in I$, given by $\chi_p(x) = 0$ if $x \neq p$, and $\chi_p(p) = 1$. The subspace topology on $Y$ is discrete, so the only dense subspace of $Y$ is $Y$ itself. Since $Y$ is also uncountable, it is not separable.

Having digested the example above, or some alternative, one understands that the statement to be proved must utilize, in a key way, the fact that the $X$ in the problem is a metric space. One way to proceed is to establish, and combine, following standard facts:

Fact 1: A separable metric space is second-countable.
Fact 2: A subspace of a second-countable topological space is second countable. In other words, second-countability is a hereditary property.
Fact 3: A second-countable topological space is separable.

Proofs of the above facts can be found in any introduction to point-set topology. However, the best course of action would be to attempt to prove them yourself first.
A: Thank you for all the responses, here I've managed to put together an answer:
Since $M \subset X$ is a metric subspace, its in its own a metric space, so it is enough to proove that $M$ is second countable, that is, that it has a countable base.
Let $\mathcal{B}$ be the countable base of $X$. Define $\mathcal{B}'=\{B\cap M:B \in \mathcal{B}\}$. Let's see that $\mathcal{B}'$ is a countable base of $M$. Let $A \subset M$ be open (in $M$). Then $A=A'\cap M$ with $A'$ open in $X$. Then $A'=\bigcup_{\lambda \in L}B_{\lambda}$ with $L$ a countable index set and $B_{\lambda}\in \mathcal{B}$ for every $\lambda$. Then
$$A=A'\cap M=\left(\bigcup_{\lambda \in L}B_{\lambda}\right)\cap M=\bigcup_{\lambda \in L}(B_{\lambda}\cap M)$$
Which implies $\mathcal{B}'$ is a base of $M$
