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 $M$?
Best Answer
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.
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:
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.