Let $q_{1},q_{2},\cdots$ of the rational in $[0,1],$ and consider the set $S:= \bigcup_{k\in\mathbb{N}} B_{r(k)}\big(q_{k}\big)$ with $r(k) = \frac{1}{2^{N}}ยท\big(\frac{1}{2}\big)^k,$ N is an integer $\geq 3. $ Why the boundary of $S$ has to be $[0,1]\setminus{S}$?
We know $S$ is an open subset of $\mathbb{R},\overline{S}=S^{ \circ}\cup\partial S=S\cup\partial S,S\cap\partial S= \emptyset.$
By definition, $\partial S = \overline S \setminus S^o$. As $S$ is dense in $[0,1]$, $\overline S = [0,1]$. Moreover, as $S$ is open, $S^o=S$. So, $\partial S = [0,1]\setminus S$
Every point in $[0,1]\setminus{S}$ is arbitrarily close to a point in $\Bbb Q$, and thus arbitrarily close to $S$, which means that it is on the boundary. On the other hand, $S$ is a union of open intervals, so it is an open set, and therefore no element in $S$ is part of the boundary.