The book by Michael Searcoid gave a different proof so I want to know where I made an error.
Proof: Let $(X,d)$ be a metric space and $S\subset X$. By definition, \begin{align} \partial S:=\{x\in X\quad\vert\quad dist(x,S)=0=dist(x,S^c)\}\\ \partial(\partial S):={y\in X\quad\vert\quad dist(y,\partial S)=0=dist(y,(\partial S)^c)} \end{align} Suppose $\alpha\in\partial(\partial S)$, then $dist(\alpha,\partial S)=0=dist(\alpha, (\partial S)^c)$; that is \begin{align} \exists\sigma\in\partial S\quad s.t.\quad d(\alpha,\sigma)=inf\{d(\alpha,\overline{\sigma})\quad\vert\quad\overline{\sigma}\in\partial S\}=0\\ \exists\sigma'\in(\partial S)^c\quad s.t.\quad d(\alpha,\sigma')=inf\{d(\alpha, \sigma')\quad\vert\quad \sigma'\in(\partial S)^c\}=0 \end{align} But, $d(\alpha,\sigma)=0=d(\alpha,\sigma')$ iff $\alpha=\sigma$ and $\alpha=\sigma'$. Therefore, $\sigma=\sigma'=\alpha\in\partial S$.
When you wrote $\alpha\in\partial S$, I suppose that you meant that $\alpha\in\partial(\partial S)$.
The error lies in assuming that if the distance of a point $\alpha$ to a set $A$ is $0$, then $\alpha\in A$. This is not true. For instance, in $\mathbb R$, with the usual distance, $d\bigl(0,(0,+\infty)\bigr)=0$, but $0\notin(0,+\infty)$.