I think this solution works. If anyone can offer corrections/improvements, I will be very thankful! We can use induction on $n$. Consider the intersection of the $n-1$-th horizontal line with the $n-1$-th vertical line; it is given that this intersection lies on the diagonal of the square. As this intersection moves along the diagonal, two squares on the main diagonal (which are known to be colored black) and several rectangles vary. The sum of the areas of the two main diagonal black squares is constant (because the length of diagonal contained between the $n-2$-th horizontal and vertical lines is constant), and it is easy to see that the area colored black is minimized either when the intersection of the $n-1$-th horizontal line with the $n-1$-th vertical line coincides with the vertex of the square or with the intersection of the $n-2$-th horizontal line with the $n-2$-th vertical line, so it suffices to prove the result for these two configurations. However, both configurations reduce to the inductive hypothesis for $n-1$ and $n-2$, respectively.