If $(A\times B) \cap (C\times D)=\varnothing$ then either $A\cap C=\varnothing$ or $B\cap D=\varnothing$. Prove that for any sets A, B, C, and D, if A × B and C × D are disjoint,
then either A and C are disjoint or B and D are disjoint.
Proof(someones).
Suppose (A X B) and (C X D) are disjoint. Let (x,y) be an arbitrary ordered pair of (A X B), it follows that $(x,y) \notin (C X D)$. So either $x \notin C$ or $y \notin D$. Since x,y are arbitrary, Thus either A and C are disjoint or B and D are disjoint.
I think the above proof is wrong since it assumes (x,y) is an arbitrary ordered pair of (A X B) without any logical justification(no existential instantination).
My Proof.(Contrapositive)
Suppose $A\cap C \ne \emptyset $ and $B\cap D \ne \emptyset$. It follows that there exist an element $a\in A\cap C$ and $b\in B\cap D$. Since $a\in A$ and $b\in B$ then $(a,b)\in A× B$ and since $a\in C$ and $b\in D$ then $(a,b)\in C×D$. So $(a,b)\in A×B \cap C×D$. So A × B and C × D not disjoint.
As for the first proof I can understand the "Suppose (A X B) and (C X D) are disjoint" part. But I cant understand the logical justification of assuming an element (x,y) in AXB since by making this assumption you are also assuming that A or B are not $\emptyset$ which makes the proof invalid as you are assuming something not given. Also A,B,C,D are supposed to be arbitrary sets.
My questions here are:
1) Is the first one correct ? If it is what is its logical justification.
2)Is my proof correct ? If not, what is my mistake ?

Your proof is correct.  The first proof is incorrect, but not really for the reason you state.  There is no harm in supposing $(x,y)\in A\times B$, because the argument is trying to make a statement for all $(x,y)\in A\times B$ (and if $A\times B$ is empty, that statement is just vacuously true).
However, the statement the first proof is proving is not the statement you are trying to prove.  It shows that if $x\in A$ and $y\in B$, then either $x\not\in C$ or $y\not\in D$.  This does not immediately imply that $A\cap C=\emptyset$ or $B\cap D=\emptyset$.  Indeed, it only proves that for all $x$ and $y$, either $x\not\in A\cap C$ or $y\not\in B\cap D$.  What you want to prove is that either (for all $x$, $x\not\in A \cap D$) or (for all $y$, $y\not\in B\cap D$).  This is a different statement, since we have moved the quantifiers inside the disjunction.
