I remember Iura's idea on this problem. Consider the leftmost square $X$. Let $A$ and $B$ are two right its vertices. Then all squares can be divided into three parts: squares containing $A$, squares containing $B$ and square nonintersecting with $X$. For squares from thord part we know that from any $k$ of them there are two intersecting squres. Therefore, by induction, third part can be partitioned into $2k-3$ subsets with required property. That's all.

yes, thats also my solution