I am trying to determine the relationship between $(A \times B) \cup (C \times D)$ and $(A \cup C) \times (B \cup D)$. Usually, I begin these proofs with a venn diagram and then formally prove the result thereafter, but there really isn't a good way that I know of to sketch a venn diagram for a Cartesian product. Some help on the intuition would be appreciated.
The first inclusion does make sense intuitively. The element on the left-hand side is of the form $(x,y)$ where $x$ is in either in $A$ or $C$ and $y$ is in either $B$ or $D$. That inclusion seems rather naturally, and the proof isn't too difficult.
Let $x \in (A \times B) \cup (C \times D)$. Then $x \in A \times B$ or $x \in C \times D$. If $x \in A \times B$, $x = (a,b)$ for some $a \in A$ and $b \in B$. Since $a \in A$, $a \in A \cup C$. Since $b \in B$, $b \in B \cup D$, so $x \in (A \cup C) \times (B \cup D)$. Alternatively, if $x \in C \times D$, then $x = (c,d)$ for some $c \in C$ and $d \in D$. Since $c \in C$, $c \in A \cup C$. Since $d \in D$, $d \in B \cup D$. Hence, $x \in (A \cup C) \times (B \cup D)$.
I have no intuition for the other side. My attempt at a proof of it (below) failed, but I have no intuition to construct a counterexample.
Let $x \in (A \cup C) \times (B \cup D)$. Then $x = (\alpha, \beta)$ where $\alpha \in A \cup C$ and $\beta \in B \cup D$. Then $\alpha \in A$ or $\alpha \in C$ and $\beta \in B$ or $\beta \in D$.
It seems that there is no guarantee that $x$ lives in the right-hand side, depending on how we "mix and match" $\alpha$ and $\beta$. With that said, I am not completely satisfied with this as an intuitive explanation. An alternative to a venn diagram (or, as someone on here recently suggested, a truth table of sorts) would be helpful.
Taking your approach for the second part $(p,q) \in (A \cup C) \times (B \cup D)$, we get $p \in A \cup C$ and $q \in B \cup D$. Thus $p \in A$ or $p \in C$ or both and a similar statement is true for $q$.
Now we want to check if $(p,q)$ necessarily lies in $(A \times B) \cup (C \times D)$. For this to happen $(p,q)$ must be in at least one of $A \times B$ or $C \times D$. That means we should have either $p \in A \quad $ AND $\quad q \in B \quad $ OR $\quad p \in C \quad $ AND $\quad q \in D$ or both.
Ask yourself:
What if $p \in A \setminus C$ and $q \in D \setminus B$? Then our initial condition (see first paragraph) is met but such $(p,q)$ will not be in $(A \times B) \times (C \cup D)$. This gives us a clue to find a possible counterexample for this containment.
Counterexample
Let $A=\{1,2\}$, $B=\{1,4\}$, $C=\{x\}$ and $D=\{y\}$. Then $1 \in A \setminus C$ and $y \in D \setminus B$ so $(1,y) \in (A \cup C) \times (B \cup D)$ but $(1,y) \not\in (A \times B) \cup (C \times D)$