Pick $a, b \in \{\text{even}, \text{odd}\}$. Take all squares with lower-left corner $(m,n)$ where $m$ has parity $a$ and $n$ has parity $b$. Clearly no two squares picked this way intersect. Each square must be chosen for some $a,b$. Since there are 4 possibilities for the pair $(a,b)$, by Pigeonhole Principle some pair $(a,b)$ chooses at least $\frac{1}{4}$ of all squares. This is clearly tight; take all squares satisfying $1 \le m,n \le 2k$ for any positive integer $k$ and see that there are only $k^2$ points in the form $(p,q)$ where $2 \le p,q \le 2k$. Each square covers one of these points, so you can take at most $k^2$ squares.