For arbitrary sets $A$ and $B$, assume the opposite of our conclusion: $A\cup B \neq B \cup A$.
Thus, either there exists $x\in (A \cup B)$ such that $x \notin (B\cup A)$ or there exists $x'\in (B\cup A)$ such that $x' \notin (A\cup B)$. Without loss of generality, assume the first condition, that there exists $x \in (A\cup B)$ such that $x \notin (B\cup A)$. Then $x\notin B$ and $x\notin A$, or in other words, $x\notin A$ and $x\notin B$. So, $x\notin (A\cup B)$, a contradiction.
From this antilogy, it follows that $A\cup B = B\cup A$.
This works, but I recommend a direct proof instead: for all $x$,
$$\begin{align} x\in A\cup B&\iff x\in A\lor x\in B\\ &\iff x\in B\lor x\in A\\ &\iff x\in B\cup A. \end{align}$$