Cross Junction

If we want to choose 2 squares from these 9 squares without any restriction ,

It can be done by $\large {9 \choose 2}$ = $\large \dfrac{9!}{7! \times 2!}$ = $\large \dfrac{9 \times 8}{2}$ = $\large 36$ ways.

But now we will use complementation. We will subtract the ways we don't want the squares to be selected from the total.

Consider the vertical column. If we consider the squares sharing an edge as 1 unit , then there are 4 pairs of squares in vertical column from which 1 can be chosen in $\large {4 \choose 1} = 4$ ways. $\large (A,B) , (B,E) , (E,H) , (H,I)$

Similarly , also from horizontal row $\large ={4 \choose 1} = 4$ ways. $\large (C,D) , (D,E) , (E,F) , (F,G)$

Hence , total ways $\large = 36 - 4 - 4 = \boxed {28}$

Start at the first square of any row/column and count the squares that don't share edges with it, it's 7.

So: $4 * 7 = 28.$