Prove that for any sets A and B there is a unique set C such that A $\Delta$ C = B
Existence: Suppose A and B are arbitrary sets. Suppose C = A $\Delta$ B. Then A $\Delta$ C = A $\Delta$ (A $\Delta$ B) = (A $\Delta$ A) $\Delta$B = $\varnothing$ $\Delta$ B = B.
Uniqueness: Suppose C$_1$ and C$_2$ are two arbitrary sets such that A $\Delta$ C$_1$ = B and A $\Delta$ C$_2$ = B. Thus A $\Delta$ C$_1$ = A $\Delta$ C$_2$. Suppose x$\in$ C$_1$. Thus we have two cases.
Case 1: x $\in$ C$_1$\A. Thus x $\in$ A $\Delta$ C$_1$ and x $\in$ A $\Delta$ C$_2$. Since x $\notin$ A, x $\in$ C$_2$.
Case 2: x $\in$ C$_1$ $\land$ x $\in$ A. Thus x $\notin$ C$_1$ $\Delta$ A and x $\notin$ C$_2$ $\Delta$ A. Since x $\in$ A, x $\in$ C$_2$.
Thus, C$_1$ $\subseteq$ C$_2$. Similar reasoning can be used to show C$_2$ $\subseteq$ C$_1$ and therefore C$_1$ = C$_2$. Therefore C is unique.
Is my reasoning for uniqueness valid? I have seen some suppose some set, say C', and let A = C' $\Delta$ B. Thus A $\Delta$ B = (C' $\Delta$ B) $\Delta$ B = C' (B $\Delta$ B) = C $\Delta$ $\varnothing$ = C. Thus C' = A $\Delta$ B = C. This seems unreasonable as A has already been assumed to be arbitrary and we are now stating it to be a particular set. The uniqueness proof comes after assuming A and B.
Your argument is fine, though like Batominovski (in the comments) I would prove uniqueness after assuming that $A\triangle C_1=A\triangle C_2$ by the calculation
$$\begin{align*} C_1&=\varnothing\triangle C_1=(A\triangle A)\triangle C_1=A\triangle(A\triangle C_1)\\ &=A\triangle(A\triangle C_2)=(A\triangle A)\triangle C_2=\varnothing\triangle C_2=C_2 \end{align*}$$
rather than by ‘element-chasing’.
The argument that you discuss in your last paragraph is a very poorly stated version of a perfectly legitimate argument that goes like this. Suppose that $A\triangle C=B$ and $A\triangle C'=B$; then
$$A\triangle B=A\triangle (A\triangle C')=(A\triangle A)\triangle C'=\varnothing\triangle C'=C'$$
and
$$A\triangle B=A\triangle (A\triangle C)=(A\triangle A)\triangle C=\varnothing\triangle C=C\;,$$
so $C'=A\triangle B=C$, and $C$ is unique.