Every closed set is a boundary [duplicate]
Can you verify the following two facts?
(i) The set $A$ as you defined has empty interior.
(ii) The closure of the set $\mathbb{Q}^n \cap \mathrm{int}(S)$ is the closure of $\mathrm{int}(S)$.
Then you can conclude as follows: $$ \partial A = \bar{A} \backslash \mathrm{int}(A) = \bar{A} = \overline{\partial S \cup (\mathbb{Q}^n \cap \mathrm{int}(S))} = \overline{\partial S} \cup \overline{\mathbb{Q}^n \cap \mathrm{int}(S)} = \partial S \cup \overline{\mathrm{int}(S)} = S. $$
If you'd like to see a proof of the two facts, I'll gladly add them.