This is related to another question, please see for the background of the problem:
Cartesian product counterexample
Part II) Find the mistake in the following bogus proof of the False theorem and fix it to show that $R \subseteq L$
Proof: Since L and R are both sets of pairs, it's sufficient to prove that $(x,y) \in L \iff (x,y) \in R$ for all $x,y$
The proof will be a chain of iff implications: $$(x,y) \in R$$ $$\iff (x,y) \in (A \times C) \cup (B \times D)$$ $$\iff (x,y) \in A \times C \text{, or } (x,y) \in B \times D $$ $$\iff (x \in A \text { and } y \in C) \text { or else } (x \in B \text { and } y \in D) $$ $$\iff \text{either } x \in A \text{ or } x \in B, \text { and either } y \in C \text{ or } y \in D$$ $$\iff x \in A \cup B \text { and } y \in C \cup D$$ $$\iff (x, y) \in L$$