From How to Prove It
Prove that for any sets A and B there is a unique set C such that $A \triangle C=B$
The immediate answer is $C=A\triangle B$, which then works, since $A \triangle (A \triangle B) = (A \triangle A) \triangle B$ by associativity of symmetric difference.
But I don’t think this answer is unique.
Notice that $A\triangle B=\{x|x\in (A-B) \vee x\in (B-A) \} = (A-B)\cup(B-A)$
Let $C=A\cup B$. Then we have $A \triangle (A \cup B) = (A-(A\cup B)) \cup (A\cup B-A)$
$(A-(A\cup B)) = \emptyset$ , since there is no x such that $x\in A$ but $x \notin (A\cup B)$.
$(A\cup B - (A)) =B$, since if $x\in A\cup B$ but $x\notin A$, it must be that $x\in B$
so $A \triangle (A \cup B) = (A-(A\cup B)) \cup (A\cup B-A)$ = B.
so C is not unique at all. But the textbook problem assumes that it is, and asks us to prove it. Is my proof that C isn’t unique incorrect? What am I missing?
The example you give is incorrect, $C = A\cup B$ does not satisfy $A\triangle C = B$. There is an equivalent definition of $\triangle$, $A\triangle B = A\cup B - A\cap B$. Then the second choice of $C = A\cup B$ you suggested gives $$A\triangle (A\cup B) = (A\cup A \cup B) -A\cap (A\cup B) = A \cup B - A\cup (A\cap B) = B- (A\cap B),$$ which is $B-A$. And $B-A$ does not contain all of $B$ when $A\cap B$ is non-trivial.
Your argument turns awry when you claim $A\cup B - A = B$. The left hand side is missing elements of $B\cap A$, which is nonempty in general, while the right hand side still has them.
The choice you first provide for $C$ is forced, if we notice that $A\triangle C = B$ means that $A \triangle (A\triangle C) = A\triangle B$. The left hand side becomes $(A\triangle A) \triangle C= \emptyset \triangle C = C$, using the facts that $S\triangle S = \emptyset$ and $\emptyset \triangle S = S$ for all $S$. Then $C = A\triangle B$.