Let the convex hull of these $(n+1)^{2}$ points be a $k$-sided polygon.
(1) If $k \geqslant 4 n$, then the perimeter of the $k$-sided polygon does not exceed $4 n$ (since the convex hull is inside the square). Therefore, there exist two consecutive sides whose lengths sum to no more than 2. Then the area of the triangle formed by these two vertices is greater than $\frac{1}{2}$.
(2) If $k < 4 n$, the convex hull divides the square into $k$ triangles.
Since the total area of the square is 1, the average area of these triangles is $\frac{1}{k}$. Given that $k < 4 n$, the average area of these triangles is greater than $\frac{1}{4 n}$. Therefore, there must be at least one triangle with an area no greater than $\frac{1}{2}$.
Hence, in either case, the original proposition holds.