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Let $a1 \in A$ and $a1  \notin B$, so then,

$(a1,a1)\notin (A \times B)\cup(B \times A)$, but $(a1,a1) \in (A \cup B) \times (A \cup B)$.

So we can clearly say that (i) is false. and you have caught the error in the proof correctly.👍

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