Proof of The Boundary of a Subset of a Metric Space Is Closed and Borel

The complement of the boundary. By Closure and Interior in a Metric Space: Containments, the Closure is Closed, the Interior is Open (i), $\operatorname{int}A\subseteq A\subseteq S$, and $\partial A=\operatorname{cl}A\setminus\operatorname{int}A\subseteq S$. We claim

$$S\setminus\partial A=(S\setminus\operatorname{cl}A)\cup\operatorname{int}A .$$

Let $z\in S$. If $z\in\operatorname{int}A$, then $z\notin\partial A$, and $z$ lies in both sides. If $z\notin\operatorname{int}A$, then $z\notin\partial A$ holds exactly when $z\notin\operatorname{cl}A$, so again $z$ lies in the left side exactly when it lies in the right side. Both sides are contained in $S$, since $\operatorname{int}A\subseteq S$, which proves the claim.

Both pieces are open. The closure $\operatorname{cl}A$ is closed in $(S,\mathcal{T}_{d})$ by Closure and Interior in a Metric Space: Containments, the Closure is Closed, the Interior is Open (ii), that is, $S\setminus\operatorname{cl}A\in\mathcal{T}_{d}$ (Closed Subset of a Topological Space (i)). The interior $\operatorname{int}A$ is open in $(S,d)$ by Closure and Interior in a Metric Space: Containments, the Closure is Closed, the Interior is Open (iii), so $\operatorname{int}A\in\mathcal{T}_{d}$.

Conclusion. By The Open Subsets of a Metric Space Form a Topology (i), $\mathcal{T}_{d}$ is a topology on $S$, so it contains the union of any two of its members (Topological Space (iii)). Hence $S\setminus\partial A\in\mathcal{T}_{d}$, and since $\partial A\subseteq S$ this says that $\partial A$ is closed in $(S,\mathcal{T}_{d})$. Finally $\partial A\in\mathcal{B}(S)$ by Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces (ii).