How do we prove that $\mu (B) = 0$ for any ball $B$ such that $B \subset S^c$?
For any $x \in S^c$, there exists $r_x > 0$ such that $\mu \left( B(x, r_x) \right) = 0$.
Why there exists a countable cover for $S^c$ such that
$$ S^c = \cup_{i = 1}^{\infty} B(p_i, r_{p_i}) \text{ and } \mu \left( B(p_i, r_{p_i}) \right) = 0 \text{ for any } i \ge 1 $$
Take $(p_i,r_i)_i$ to be an enumeration of $A = \{r,q \in \mathbb Q\times \mathbb Q^n : \mu(B(p,r))=0\}$.
Then, if $x \in S^c$, then there is $r>0$ such that $\mu(B(x,r)) = 0$. Now we just need to find $p,r'\in \mathbb Q^n \times \mathbb Q$ such that $$x\in B(p,r') \subset B(x,r).$$
This is easy, by density of rationals, and a standard drawing of a picture of a ball inside another ball.