If $\mathbb{R}=\bigcup_{n=1}^{\infty}E_n$, then the closure of some $E_n$ contains an interval I'm working on some problems in Carothers' Real Analysis. I just started the section on the Baire Category Theorem. Thus far Carothers has given the Baire Category Theorem for $\mathbb{R}$.

Prove that if $\mathbb{R}=\bigcup_{n=1}^{\infty}E_n$, then the closure of some $E_n$ contains an interval. That is, $\operatorname{int}(\overline{E}_n)\neq \emptyset$, for some $n$.

Thoughts/Attempt:
Not really sure where to start on this one, so I'll just state some stuff I know.
The Baire theorem we learned for $\mathbb{R}$ states that the infinite intersection of open dense sets in $\mathbb{R}$ is non-empty, and is in fact dense in $\mathbb{R}$ as well. An interval $(a,b)\subset \mathbb{R}$ has limit points $a$ and $b$. A hint would be appreciated to get started. Thank you.

If $\mathbb{R} = \cup_{n=1}^\infty E_n$ where int$(\overline{E}) = \emptyset$ for all $n$ then $\mathbb{R}$ is a countable union of nowhere dense sets contradicting the Baire Category Theorem.
