Let $A$, $B$, $C$ be points in $S$ such that $[ABC]$ is maximized. Let $\ell_1$ be the line passing through $A$ which is parallel to $BC$. If there exists a point $X \in S$ which lies on a different side of $\ell_1$ as $B$, then the distance from $X$ to $BC$ is greater than that from $A$ to $BC$. This yields the contradiction $[XBC] > [ABC]$. Therefore, $X$ must lie on the same side of $\ell_1$ as $B$.

Similarly, let $\ell_2$ be the line passing through $B$ which is parallel to $CA$, and let $\ell_3$ be the line passing through $C$ which is parallel to $AB$. Then any point in $S$ must lie inside the triangle $\triangle DEF$ bounded by $\ell_1$, $\ell_2$, $\ell_3$. Note that $\triangle ABC$ is the median triangle of $\triangle DEF$. So $[DEF] = 4[ABC] < 4$. Thus, we can simply take $\triangle DEF$.