Consider the fact $$ (A \subseteq B) \land (C \subseteq D) \iff (A \times C) \subseteq (B \times D) $$ where A,B,C and D are sets.
Rewrite that as operations on elements of given sets: $$ \forall x : (x \in A \Rightarrow x \in B) \land \forall y : (y \in C \Rightarrow y \in D) \iff (\{\langle x,y \rangle \mid x \in A \land y \in C\} \subseteq \{ \langle x,y \rangle \mid x \in B \land y \in D \}) $$
Rwrite the right part using the definition of subset and move quantifiers in the left part: $$ \forall x \forall y : ((x \in A \Rightarrow x \in B) \land (y \in C \Rightarrow y \in D)) \iff \\ \forall x \forall y : ((x \in A \land y \in C) \Rightarrow (x \in B \land y \in D)) $$
Move the quantifiers to the beginning once again: $$ \forall x \forall y : ((x \in A \Rightarrow x \in B) \land (y \in C \Rightarrow y \in D)) \iff \\ ((x \in A \land y \in C) \Rightarrow (x \in B \land y \in D)) $$
Now replace the "belongs" statements with predicates: $$ A' := (x \in A) \\ B' := (x \in B) \\ C' := (y \in C) \\ D' := (y \in D) \\ $$ Thus we get an equivalent formula: $$ ((A' \Rightarrow B') \land (C' \Rightarrow D')) \iff ((A' \land C') \Rightarrow (B' \land D')) $$ The last statement in terms of logic is not correct (e.g. assign $A':=1; B':=0; C':=0; D':=1)$, yet the initial set theoretical statement is correct. This suggests that at least one of the transformations above is wrong. Could it be a mistake in the logic itself or the transition between set theoretical notation and logical statements? I can not see where exactly the transformations went wrong. The implication still does hold, that is, the left part is still sufficient for the right part to hold, but it is not a necessary condition.
Note: the original statement is not (necessarily) correct: if $A = \emptyset$ and $C \not\subseteq D$, we still have $A \times C \subseteq B \times D$.
But really the problem occurs where you move the $\forall x\forall y$ outside; generally it is not true that $$ \forall x(\phi(x) \leftrightarrow \psi(x)) \iff \forall x\phi(x) \leftrightarrow\forall x \psi(x) $$ and it should not be difficult to come up with examples. (In fact, you already have.)