A nowhere dense subset of $\mathbb{R}$ is countable

Question

Let $E$ be a nowhere dense subset of $\mathbb R$. How would I show that $E$ is countable?

Since $E$ is nowhere dense, $\overline E$ contains no nonempty open subset of $\mathbb R$. Let $x\in\overline E$. Then, for any $\varepsilon>0$, we have $(x-\varepsilon,x+\varepsilon)\not\subset\overline E$.

That's all I can come up with. I'm sure that I need to show that $E$ is the countable union of singletons, but I'm not sure how to arrive that this result.

Answer 1

The Cantor set is uncountable and nowhere dense.

Answer 2

Take any open set $U$ containing $\mathbb Q$ of finite measure. Then $\mathbb R \setminus U$ is closed and nowhere dense, and since it has positive measure, it is uncountable.