# Countable Basis Proof in Munkres

Munkres writes the following:

Suppose that $$X$$ has a countable basis. Then there exists a countable subset of $$X$$ that is dense in $$X$$.

The proof:

From each nonempty basis element $$B_n$$, choose a point $$x_n$$. Let $$D$$ be the set consisting of the points $$x_n$$. Then $$D$$ is dense in $$X$$: Given any point $$x \in X$$, every basis element containing $$X$$ intersects $$D$$, so $$x$$ belongs to $$\overline{D}$$.

I don't understand why the bold statement is true. Why can't an open set containing $$x$$ have empty intersection with $$D$$?

If $$x\in X$$, every basis element $$U$$ containing $$x$$ is equal to $$B_n$$, for some natural $$n$$. But then $$x_n\in B_n=U$$. Since $$x_n\in D$$, this proves that $$U\cap D\neq\emptyset$$. Since every non-empty open set contains some basis element, this proves that every non-empty open set intersects $$D$$. In other words, $$D$$ is dense in $$X$$.