$A \subset X$ is nowhere dense in $X$ IFF each non-empty open set in $X$ contains an open ball whose closure is disjoint from $A$.

I want to prove that: a subset $$A$$ of a metric space $$X$$ is nowhere dense in $$X$$ IFF each non-empty open set in $$X$$ contains an open ball whose closure is disjoint from $$A$$.

This is a lemma that is used in Baire's category theory.

Definitions:

• $$A \subset X$$ is nowhere dense if $$\text{int}\left(\bar{A}\right) = \emptyset$$.
• A topologial space is called $$T_1$$ if for each pair $$a$$, $$b$$ of distinct points of $$X$$, there are open set $$U$$ and $$V$$ in $$X$$ such that $$a \in U, a \not \in V, b \not \in U$$ and $$b \in V$$.
• A topologial space is called regular if it is a $$T_1$$ space and for each closed subset $$C$$ and each point $$a$$ not in $$C$$, there exist disjoint open sets $$U$$ and $$V$$ in $$X$$ such that $$a \in U$$ and $$C \subset V$$.

My proof so far:

$$\Rightarrow:$$

Assume that $$A \subset X$$ is nowhere dense. If $$X$$ is empty, then the statement trivially holds. So assume that $$X$$ is not empty. If $$\bar{A} = X$$ then $$\text{int}\left( \bar{A} \right) = X \not = \emptyset$$, which contradicts the assumption that $$A$$ is nowhere dense. So we have that $$\bar{A} \not = X$$. So there exists an $$a$$ not in $$\bar{A}$$. Since any metric space is regular and $$\bar{A}$$ is closed we obtain an open ball $$B_a(r)$$ such that $$B_a(r)$$ and $$\bar{A}$$ are disjoint. But then also $$B_a(r)$$ and $$A$$ are disjoint. Now we note that $$\overline{B_a(r/2)} \subset B_a(r)$$. So we find that $$\overline{B_a(r/2)}$$ is disjoint from $$A$$.

The other direction I don't fully have. I made a start.

$$\Leftarrow:$$

Assume that each non-empty open set in $$X$$ contains an open ball whose closure is disjoint from $$A$$. And for a contradiction assume that $$A$$ is not nowhere dense. Then $$\text{int}\left( \bar{A} \right) \not = \emptyset$$. So by the assumption there exists an open ball $$B$$ with $${B} \subset \text{int}\left( \bar{A} \right)$$ and $$\overline{B} \cap A = \emptyset$$. (Here I'm stuck)

It seems to me that we can derive a contradiction, but I don't know how.

I would like to have help with that and also receive comments on the rest of my proof.

Write $$B=B(x,r)$$ since $$B\subset \bar A$$ there exists a sequence $$x_n$$ in $$A$$ which converges towards $$x$$, this implies by definition there exists $$N$$ such that $$n>N$$ implies that $$x_n\in B$$. Contradiction since $$B\cap A\subset \bar B\cap A$$ is empty.