For every set $S \subseteq \mathbf{R}$, show that $int(S) \subseteq ext(ext(S))$. I would like to ask, if the following constitutes a valid proof, or should I attempt it again?
Let $ext(S)$ denote exterior of set $S \subseteq \mathbb R$.

For every set $S \subseteq \mathbf{R}$, show that $int(S) \subseteq ext(ext(S))$.

My attempt.
Let $x$ be an arbitrary point in $int(S)$. By definition of the interior of a set, there exists an $\epsilon > 0$ such that the open interval $(x - \epsilon,  x + \epsilon)$ is contained in $S$. We are interested to show that, $(x - \epsilon, x + \epsilon) \in (ext(S))^C$.  Pick an arbitrary $y \in (x - \epsilon, x + \epsilon) \subseteq S$. We choose $\delta = \min \{\frac{y-(x-\epsilon)}{2},\frac{(x+\epsilon)-y}{2}\}$, so $(y - \delta, y+\delta) \subseteq (x - \epsilon, x + \epsilon) \not\subseteq S^C$. So, $y \notin ext(S)$. Therefore, $y \in (ext(S))^C$. This is true for all $y \in (x - \epsilon, x + \epsilon)$. So, $(x - \epsilon, x + \epsilon) \subseteq (ext(S))^C$. Consequently, $x \in (ext(ext(S))$.
Thus, $int(S) \subseteq ext(ext(S))$.

Easier : if $x \in O=\operatorname{int}(S)$ then the open $O$ witnesses also that $O$ is a subset of $\operatorname{ext}(S)^\complement$ so in its interior i.e. $\operatorname{ext}(\operatorname{ext}(S))$ and so you’re done showing the inclusion. The metric or order plays no role.
