Trying to develop a proof for the following statement. I don't think my proof is quite correct.
Statement: Let A, B be nonempty sets and let C, D be sets. If $A \times B \subseteq C \times D$, then $A \subseteq C$ and $B \subseteq D$.
Proof: Let a be any element of A and let b be any element of B. Then (a,b) $\in A \times B$. Since $A \times B \subseteq C \times D$, (a, b) $\in C \times D$. Hence, $a \in C$ and $b \in D$. Therefore, $A \subseteq C$ and $B \subseteq D$.
Best Answer
Almost, but there is something very subtle going on that makes your proof not quite correct. You need to prove two separate statements: $A \subseteq C$ and $B \subseteq D$. Somewhere you must use the hypothesis that $A$ and $B$ are nonempty; otherwise this is not necessarily true! (For example: $\emptyset \times B \subseteq \emptyset \times \emptyset$ for any set $B$.)
So to prove $A \subseteq C$, we can work like this: Suppose $a \in A$. Then since $B$ is nonempty, there exists some $b \in B$. Now you can use your argument with $(a,b)$ to show that $a \in C$. Similarly you can prove that $B \subseteq D$, this time using that $A$ is nonempty, but this is a different argument (because you must obtain $a$ and $b$ differently).
To hopefully make it clear, when proving that $A \subseteq C$ (i.e. $\forall a. a \in A \Rightarrow a \in C$) you can assume $a \in A$ is given, but you must explain why $b \in B$ exists before you can use it.
Let me put it one more way: Proving $(\forall a. a \in A \Rightarrow a \in C) \wedge (\forall b. b \in B \Rightarrow d \in D)$ is different from proving $\forall a. \forall b. (a \in A \wedge b \in B) \Rightarrow (a \in C \wedge b \in D)$. You proved the latter, but you were asked to prove the former.
(I can imagine why a lot of people slip up on this one.)