Set theory (cartesian products) - What is wrong with this proof?

(A × B) $\cup$ (C × D) $\subseteq$ (A $\cup$ C) × (B $\cup$ D)

Notice that above we do not have equality, but containment. What is wrong
with the following attempt at proving the containment the other way round?
Incorrect Theorem. For sets A, B, C, and D we have (A $\cup$ C) × (B $\cup$ D) $\subseteq$ (A × B) $\cup$ (C × D).
“Proof”. Suppose (x, y) $\in$ (A $\cup$ C) × (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 two cases
separately.
Case 1. x $\in$ A and y $\in$ B. Then (x, y) $\in$ A × B.
Case 2. x $\in$ C and y $\in$ D. Then (x, y) $\in$ C × D.
Thus either (x, y) $\in$ (A×B) or (x, y) $\in$ C ×D, so (x, y) $\in$ (A×B)$\cup$(C ×D).

... see the thread you posted with the original question

...

Neg?