Is it true , if $|A|=|B|$ and $|C|=|D|$, then $|A \times C| = |B \times D|$? Check my proof, please. 
Divide into subsets $A \times C$ and $B \times D$ so that , all pairs with the same element belong to the same subset. Each such subset $|A \times C|$ bijective $C$, $|C|=|D|$ and $D$ equipotent to each subset $B \times D$. This implies , that $|A \times C| = |B \times D|$.
 A: Your suggestion is not a proof. At all. It is a description of a possible proof, and a hand-wavy too.
It doesn't explain why you can make these divisions, and why in making them you can still make this translation from $A\times C$ to $B\times D$.

Instead, you have to start by using what you're given. You are given that $|A|=|B|$ and $|C|=|D|$. Namely there are functions $f\colon A\to B$ and $g\colon C\to D$ which are bijections.
Now you have to use this information to show that there is a bijection between $A\times C$ and $B\times D$.
Recall that the elements of $A\times C$ are ordered pairs $(a,c)$ such that $a\in A$ and $c\in C$, and similarly the elements of $B\times D$ are ordered pairs $(b,d)$ such that $b\in B$ and $d\in D$.
Look at the information we inferred from the assumptions. Do you see a way of translating an ordered $(a,c)$ into an element of $B\times D$?
HINT: If we have a way of translating $a$ into $b$ and $c$ into $d$, perhaps we can "lift" those translations into a translation of ordered pairs.
