My proof (so far):

Let $(a,c)\in A\times C$, then $a\in A$ and $c\in C$.

Let $(b,d)\in B\times D$, $b\in B$ and $d\in D$.

If $p\in A\times C$ then $p=(a,c)$.

Define $(f\times g)(p)$ as $(f\times g)(p)=(f\times g)(a,c)=(f(a),g(c))=(b,d)$, however $b\in B$ and $d\in D$ and $(b,d)\in B\times D$.

This shows $f\times g$ is a function from $A\times C$ to $B\times D$ as $(f\times g):A\times C\to B\times D$ then $(a,b)\to (f(a),g(c))$.

I know that it is not well written out, but I was wondering if my thought process so far made sense or if I accidentally missed something. Additionally, I am unsure of what the next step may be. As a sidenote, I am worried that this does not work for the given definition of $f\times g$.