A stupid question: The square with side length 1is inside ABCD, does that mean it must lie strictly in ABCD or the sides of the square can can be on the sides of ABCD?

It is unclear that the square with side length $1$ has to necessarily be parallel to the sides of the big square, but in both ways the answer is $15$. Divide the big square to $16$ Squares of side length $1$. Clearly at least one of the squares is without a point. To say for $16$ it is not necessarily correct, consider this configuration: Put a point on the center of each of the $16$ small squares, then from the center of the big square, dilate with ratio $\alpha$ such that $\frac{2}{3}<\alpha <\frac{\sqrt{2}}{2}$. Consider The $16$ new points, it’s not that hard to see that there are no squares with side length $1$ (parallel or otherwise) in this configuration.