Q:

Nowhere dense set 3

Let $ M \subset X, $ where $X $ is a metric space. Prove that if $ M $ is nowhere dense(rare) in X, then the set $(X \setminus \overline{M}), $ is dense in $X$.
Actually, the converse is also true ,but I was able to prove the converse. With this side,  I tried the following,
Given $ x\in \overline{M} $, since $M$ is rare, we have $ \mathrm{Int}(\overline{M})= \emptyset $. This means, $ \forall r>0 \quad B(x,r) \cap (X \setminus \overline{M}) \ne \emptyset.$, In particular, we can take $r=1$ So, there exist an element $x_1 \in B(x,r) \cap (X \setminus \overline{M})$. Then , since $B(x,r) \cap (X \setminus \overline{M})$ is an open set, $ \exists r_1 \leq \frac{1}{2}$ s.t $ B(x_1, r_1) \in B(x,r) \cap (X \setminus \overline{M}) $.
I'm trying to construct a sequence that converges to $x$, but I could not proceed from here. Was my starting point is wrong? If not, give me a direction to finish the proof.

A:

Given $x \in X$ , we need to show that if $U\subset X$  is an open set containing $x$ ,it holds that $U \cap (X-\overline M) \ne \emptyset$.
if $x \in X-\overline M $ this is trivial.
otherwise $x \in \overline M$. $M$ is rare, so if $x \in U $ we have $U \cap(X- \overline M) \ne \emptyset$.

