@TedShifrin running into a problem in showing $S' = S \cup S^{F}$, is closed, where $S^{F}$ is the set of frontier points.

I want to do it showing the union of the complements is open: $\mathbb{R}^{n} \setminus (S \cup S^{F}) = (\mathbb{R}^{n} \setminus S) \cup (\mathbb{R}^{n} \setminus S^{F})$. I know $\mathbb{R}^{n} \setminus S^{F}$ is open because I proved $S^{F}$ is closed in the previous question.

I don't know much about $S$ except that it is composed of interior points and frontier points. Frontier points are taken care of in $S^{F}$, but I'm not sure how I can claim $\mathbb{R}…

I feel I got most of the proof right in showing the closure is the smallest closed set:

let $a \in \mathbb{R}^{n} \setminus C$ this implies $a \notin S\ \Rightarrow a \in [S^{F} \cup (\mathbb{R}^{n} \setminus S)]$. But $a \notin S^{F}$ because $S^{F}$ is closed. So $a \notin (S \cup S^{F}) = \mathbb{R}^{n} \setminus S'$.

This is on your hint of showing $\mathbb{R}^{n} \setminus C \subset \mathbb{R}^{n} \setminus S'$
The only thing I feel uneasy about is using that $S^{F}$ is closed the way that I did.