Is this short proof ambiguous? Prove $\partial \partial S \subset \partial S.$

Question

Would you consider this proof to be ambiguous and, therefore, insufficient?

Note: $\partial S$ is the set of boundary points of a subset $S \subset M$ of some metric $M.$

Prove $\partial \partial S \subset \partial S.$

Consider any point $p \in \partial \partial S.$ All neighborhoods $N_rp$ of $p$ contain points in $\partial S$ and points in $(\partial S)^c.$ Suppose $p \not\in \partial S,$ that is, $p \in (\partial S)^c.$ Then, there is some neighborhood $N_{r'}p$ that contains no points in $\partial S,$ namely the neighborhood $N_{r'}p$ where $r' < d(p,p')$ for the nearest point $p' \in \partial S$ to $p,$ contradicting the initial assumption. Hence, $p \in \partial S.$

Answer 1

In fact you can argue equally shortly using the definition of boundary in a topological space $X$: $x \in \partial E$ if and only if every neighborhood contains a point of $E$ and a point of $X \setminus E$.

If $x \in \partial \partial S$ and $U$ is a neighborhood of $x$, then $U$ contains a point $y \in \partial S$. Since $U$ is a neighborhood of $y$, $U$ contains a point of $S$ and a point of $X \setminus S$. It follows $x \in \partial S$.

Answer 2

Exercise. Prove
$\partial \partial A = \partial A -$interior $\partial A.$