What's wrong with the following proof that for any sets $A, B, C$, and $D,(A \cup C) \times(B \cup D) \subseteq(A \times B) \cup(C \times D)$ ? (Note that this is the reverse of the inclusion in part 4 of Theorem 4.1.3.)
Proof. Suppose $(x, y) \in(A \cup C) \times(B \cup D)$. Then $x \in A \cup C$ and $y \in B \cup D$, so either $x \in A$ or $x \in C$, and either $y \in B$ or $y \in D$. We consider these cases separately.
Case I. $x \in A$ and $y \in B$. Then $(x, y) \in A \times B$.
Case 2. $x \in C$ and $y \in D$. Then $(x, y) \in C \times D$.
Thus, either $(x, y) \in A \times B$ or $(x, y) \in C \times D$, so $(x, y) \in(A \times B) \cup$ $(C \times D)$.