Prove that if $A\times B$ is nonempty then there is a $C$ in $A\times B$ such the the intersection of $\bigl(\bigcup C\bigr)$ with $A\times B$ is empty.
I've been struggling with this problem for quite a long time. It is in Axiomatic Set Theory by Suppes pg. 56. Based on where it is in the book, I don't think it relies on the Axiom of Choice. My instructor hinted that it uses the axiom of regularity, but I haven't found the appropriate set to apply it to.
Working in $ZF$; for a proof by contradiction, lets suppose that every element $C$ of $A \times B$ has its union intersecting with $A,B$ in a nonempty manner. Fix some element $C$ of $A \times B$, then we can construct the intersectional set $K$ of all inductive sets $I$ such that: $$C \in I \land \forall b \forall d (b\in I \land d \in \cup(b) \cap [A \times B] \to d \in I)$$ Now take $TC(K)$ [i.e.; the transitive closure of $K$] and separate on it by formula $$y \in K \lor \exists a,b \in K (a \in y \in b)$$; and the resulting set would clearly violate foundation!