Let $D \subset \mathbb{R}$ and let $D^\circ$ be the interior of $D$. So if $x \notin D^\circ$ then $x$ must be in the closure $\mathbb{R}\setminus D$.
My attempt at this: Suppose $x \notin D^\circ$. Then $x$ is not an interior point of $D$, i.e there does not exist $\delta > 0$ such that $(x-\delta,x+\delta) \subset D$. So either $x \in D$ and $x$ is not an interior point or $x \in \mathbb{R}\setminus D$ and is an interior point of $\mathbb{R}\setminus D$ or $x \in \mathbb{R}\setminus D$ and is not an interior point of $\mathbb{R}\setminus D$.
From here it seems like I have to go case by case, but it seems unclear how this will even get closure for this from here. Also do I even need the first case, as it tells me nothing about $\mathbb{R}\setminus D$. Thanks in advance.