Consider what must happen in order for the minimum distance to be exactly 5 . Let one square, say $S_{1}$ have vertices of $(0,0),(0,1),(1,0)$, and $(1,1)$. Further, assume WLOG that the center of $S_{2}$ is above the line $y=\frac{1}{2}$ and to the right of the line $x=\frac{1}{2}$, determined by the center of $S_{1}$. There are three cases to consider: - the right side of $S_{1}$ and the left side of $S_{2}$ are 5 units apart, and the bottom left vertex of $S_{2}$ lies under the line $y=1$; - the top side of $S_{1}$ and the bottom side of $S_{2}$ are 5 units apart, and the bottom left vertex of $S_{2}$ lies to the left of the line $x=1$; - the bottom left coordinate of $S_{2}$ is $(a, b)$ with $a, b \geq 1$, and $5=\sqrt{(a-1)^{2}+(b-1)^{2}}$. We see that the first two cases are symmetric, so consider the case where the left edge of $S_{2}$ lies on the line $x=6$. When this is true, the maximum distance will be achieved between $(0,0)$ and the upper right vertex of $S_{2}$. The upper right vertex can achieve the points $(7, c)$ where $1 \leq c \leq 2$, and so $y \in[\sqrt{50}, \sqrt{53}]$. The other case we have to consider is when the bottom left vertex of $S_{2},(a, b)$, is above $y=1$ and to the right of $x=1$, in which case the maximum distance is achieved from $(0,0)$ and the upper right vertex of $S_{2}$. This distance is $\sqrt{(a+1)^{2}+(b+1)^{2}}$, which, by the triangle inequality, is at most $\sqrt{(a-1)^{2}+(b-1)^{2}}+\sqrt{2^{2}+2^{2}}=5+2 \sqrt{2}$. Since equality holds when $a=b=5 / \sqrt{2}+1$, the largest possible maximum here is $5+2 \sqrt{2}$, and the difference between the largest and smallest possible values of $y$ is $5+2 \sqrt{2}-\sqrt{50}=5-3 \sqrt{2}$.