I was trying to prove the following:
For any sets $A$ and $B$, if $(A) ∪ (B) = (A ∪ B)$ then $A ⊆ B$ or $B ⊆ A$.
My proof: Let $A$ and $B$ be arbitrary sets. Suppose $(A) ∪ (B) = (A ∪ B)$. Assume, by way of contradiction, that $(¬(A ⊆ B) ∧ ¬(B ⊆ A))$. Then there are elements $y$ and $x$, such that $x ∈ A ∧ x ∉ B$ and $y ∈ B ∧ y ∉ A$. Notice $x ∈ A ∪ B$, and $y ∈ A ∪ B$. Let $W = \lbrace x,y \rbrace$. Then $W ⊆ A ∪ B$. So $W ∈ (A ∪ B)$. Hence $W ∈ (A) ∪ (B)$. Assume $W ∈ (A)$. Then $W ⊆ A$, so $x,y ∈ A$. But $y ∉ A$, so $W ∉ (A)$. Then $W ∈ (B)$, but then $x,y ∈ B$, in particular $x ∈ B$. But that's a contradiction, so $W ∉ (b)$. Thus, we have $W ∈ (A) ∪ (B)$ and $W ∉ (A)$ and $W ∉ (B)$. Therefore, our assumption that $(¬(A ⊆ B) ∧ ¬(B ⊆ A))$ must be wrong. Ergo, $A ⊆ B$ or $B ⊆ A$.
I am suspicious of the step where I create a set $W = \lbrace x,y\rbrace$. It seems a bit odd, to be able to create a set out of the blue, especially since $W$ is not arbitrary, it includes $x$ and $y$. While I can see that $x$ and $y$ must exist, I don't see why they must be in the same set.
Is this proof correct?