Boundary operator's idempotence Quoted from Wiki

For any set $S$, $∂S ⊇ ∂∂S$, with equality holding if and only if the
  boundary of $S$ has no interior points, which will be the case for
  example if $S$ is either closed or open.

I was wondering when the boundary of $S$ has an interior point? How is it the case for a subset neither closed nor open?
Thanks and regards!
 A: If $S$ is open, then $\partial S$ is nowhere dense and closed, i.e. it has no interior points.
Also, this holds if $S$ is closed.
Proof: suppose $x \in O \subset \partial S$, where $O$ is open and $S$ is open as well.
Then, because $x \in \partial S$, $O$ must intersect $S$, say in $y$ and $y \in O \cap S$, the latter set is open and sits inside $S$, so misses $X \setminus S$, which contradicts that $y$ is in $\partial S$, which it must be (as $O \cap S \subset O \subset \partial S$), contradiction.
Suppose now that $S$ is closed and $x \in O \subset \partial S$. As $S$ is closed, $\partial S \subset S$, so $x$ would then be a boundary point of $S$ and an interior point of $S$ (as witnessed by $O$), and this cannot be.
This explains the remark "when $\partial S$ has no interior points, which will be the case for example if $S$ is either open or closed". We have seen that when $S$ is open or closed, $\partial S$ has no interior points and then we have, according to this article, that $\partial \partial S = \partial S$. 
But this is not a necessary condition: if $S = [0,1)$ in $\mathbb{R}$, then $S$ is not open, and not closed, but $\partial S = \{0, 1\}$, which equals $\partial \partial S$, so the equality does not imply that $S$ is open or closed; it can happen in other cases as well. If $S = \mathbb{Q}$, then $\partial S = \mathbb{R}$ (every non-empty open set in $\mathbb{R}$ contains rationals and irrationals), which has interior points of course, and then $\partial \partial S = \emptyset$, which shows there can be a large gap between the double boundary and the single one...
