Let $A,B,C$ and $D$ be sets such that $A\neq\emptyset$ and $B\neq\emptyset$. Show that $A\subset B$ and $C\subset D$ if, and only if, $A\times C \subset B\times D$.
I think I managed to do it correctly, but I'm not sure. Here's my proof attempt:
$(\rightarrow)$ Suppose $(x,y)\in A\times C$. $(x,y)\in A\times C \iff x\in A\;\wedge\;y\in C \implies x\in B\;\wedge\;y\in D \iff (x,y)\in B\times D$.
$(\leftarrow)$ Proof by contrapositive: We will show that $A\not\subset B \;\lor\;C\not\subset D \implies A\times C \not\subset B\times D$. Without loss of generality let's assume $A\not\subset B$. $x\in A \implies x\in A \;\land\; x\notin B$. If $C = \emptyset$ then $\emptyset \not\subset D$, which is an absurd since $\emptyset \subset M$ for all set $M$. Then let's take $y\in C$. $y\in C \implies y\notin D$. Then we have $x\in A \land x\notin B \land y\in C \land y\notin D \iff x\in A \land y\in C \land x\notin B\land y\notin D \iff (x,y)\in A\times C \land (x,y)\notin B\times D.\;$ QED
Is my proof alright? Also, is there anyway to prove the $(\leftarrow)$ part without using proof by contrapositive?