Choose two points $X$ and $Y$ from $S$, and consider the half plane $h$ with boundary the line $XY$, and containing at least one other point of $S$. Now select a point $Z$ of $S$ in the interior of $h$ whose distance from the line $XY$ is a minimum. If there are several such points, choose an arbitrary one.

Now claim that the triangle $XYZ$ does not contain any other point of $S$, apart from $X$, $Y$, and $Z$. Suppose there is a point $A \in S$ which is in the interior of the triangle $XYZ$. Then $\operatorname{area}(AXY) < \operatorname{area}(ZXY)$ and so the altitude from $A$ of the triangle $AXY$ is less than the altitude from $Z$ of the triangle $ZXY$. This contradicts the choice of $Z$.

Suppose that there are $k$ pairs $(X, Y)$ of points in $S$ with the property that the whole of the set $S$ lies in one of the two half planes with boundary the line $XY$. Such pairs are adjacent vertices of the convex hull of $S$ and so $k \le n$. For each of these pairs $(X, Y)$ we have shown that there is at least one triangle $XYZ, Z \in S$, which is “empty”, that is it does not contain any of the remaining $n-3$ points of $S$.

The remaining $\binom{n}{2} - k$ pairs $(X, Y)$ have the property that there are at least two “empty” triangles $XYZ, Z \in S$. Now sum all these triangles over all the pairs $(X, Y)$ and we get in all
$$k + 2 \left( \binom{n}{2} - k \right) = n(n-1) - k \geq n(n-1) - n = n(n-2)$$
such triangles. Each is counted at most three times so the number of “empty” triangles is at least $\frac{n(n-2)}{3}$.