Suppose not. Let the initial dimensions be $a \times b$. Let $a/b$ in reduced form be $x/y$. Fix one child. For that child, let the side length be $r$. Then the number of squares along $a$ is $a/r$ and that along $b$ is $b/r$, and their ratio is $x/y$. So $x$ divides $a/r$ and $y$ divides $b/r$. So $xy$ divides the number of squares. Summing over all children, $xy$ divides the total number of squares, and since $a \ne b$, $xy>1$, contradiction.