Consider the following theorem.
$\textbf{Theorem:}$ for any sets $A, B, C, D$, if $A \times B \subseteq C \times D$ then $A \subseteq C$ and $B \subseteq D$.
Then the following proof is given. $\textbf{Proof:}$ Suppose $A \times B \subseteq C \times D$. Let $a$ be an arbitrary element of $A$ and let $b$ be an arbitrary element of $B$. Then $(a,b) \in A\times B$. so since $A\times B \subseteq C \times D$, $(a,b) \in C \times D$. Therefore $a \in C$ and $b \in D$. Since $a$ and $b$ were arbitrary elements of $A$ and $B$, respectively, this shows that $A \subseteq C$ and $B \subseteq D$. QED.
I know that the theorem is not correct, because there is a counterexample $A = \left\lbrace 1 \right\rbrace$, $B = C = D = \emptyset$. Notice that $A \times B \subseteq C \times D \sim \emptyset \subseteq \emptyset$ but $A \nsubseteq C$. So, clearly, this is an invalid proof, but I cannot figure out which step is wrong.