How to show that the complement of Cartesian product of two non-empty sets is not the same as the Cartesian product of their complements?

$(A\times B{)}^{C}=({A}^{C}\times {B}^{C})\cup ({A}^{C}\times B)\cup (A\times {B}^{C})$

I wanted to do it like this:

$A\times B=\{(a,b)\in A\times B|a\in A\wedge b\in B\}$

Taking complement, $\overline{A\times B}=\{(a,b)\notin A\times B|a\notin A\vee b\notin B\}$

Because $(a,b)\notin A\times B$ prevents me from going anywhere, I chose to do:

Let $A\times B\subseteq X\times Y$

Then, $A\times B=\{(a,b)\in X\times Y|a\in A\wedge b\in B\}$ and $\overline{A\times B}=\{(a,b)\in X\times Y|a\notin A\vee b\notin B\}$

But I'm unsure about the correctness of what I did from here.

$\overline{A\times B}=\{(a,b)\in X\times Y|(a\notin A\wedge b\notin B)\vee (a\notin A\wedge b\in B)\vee (a\in A\wedge b\notin B)\}$

Therefore, $\overline{A\times B}=\{(a,b)\in X\times Y|(a\notin A\wedge b\notin B)\}\cup \{(a,b)\in X\times Y|(a\notin A\wedge b\in B)\}\cup \{(a,b)\in X\times Y|(a\in A\wedge b\notin B)\}$ which gives

We know $A\times B\subseteq A\times B$.

Hence, $\overline{A\times B}=(\overline{A}\times \overline{B})\cup (\overline{A}\times B)\cup (A\times \overline{B})$.

"Derive" instead of "prove" because I don't need to prove the equivalence per se. I'm just trying to see how the LHS can lead to the RHS. I should've been more careful about the wording