I did A x A and the result was [(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)]
I also tried X x Y keeping in mind that the subsets are pairs: X = [a, b]; Y = [a', b']
X x Y = [ (a,b), (a',b), (a, b'), (a',b')]
And i stated that any S which has 2 pairs as subsets must be a solution if one of those isnt (1,1), (2,2), (3,3). So S can be for example (1,2), (1,3).
Is this correct ? Thanks a lot
Your example is not correct since $S=\{1\} \times \{2,3\}$. Take $S=\{(1,2),(2,3)\}$ instead. If $S=A\times B$ then $1$ and $2$ both belong to $A$ and $2$ and $3$ belong to $B$. So we must have $(1,3) \in S$ a contradiction.