I found this exercise in my textbook, and the correction proposed tries to solve the problem via double inclusion, whereas I opted for a proof by contradiction, though I'm almost sure that I've made a mistake somewhere.
We define $A\Delta B=(A\backslash B)\cup(B\backslash A)$, where $A,B,C$ are all subsets of $E$. Is this proof that if $A\Delta B=A\Delta C$ then $B=C$ valid?
We will do a proof by contradiction. Suppose that $C\neq B$. This means that $\exists x\in C$ such that $x\in C\backslash B$ and $x\in A\Delta C\Rightarrow x\in A\Delta B$. Consequently, this means that $x\in A\backslash B$ or $B\backslash A$. Because $x\notin B$ the latter is impossible and consequently we have necessarily $x\in A\backslash B$ and so $x\in A$.
$x\in A\Delta C\Rightarrow x\in A\backslash C$ or $x\in C\backslash A$. $x\in C$ so we cannot have $x\in A\backslash C$. But this is impossible, as we have shown that $x\in C\backslash A$ and $x\in A\backslash B$. Hence this equality is only plausible if $B=C$!
There's a handful of gaps in your proof.
First, is what I mentioned in my comment but I'll put it here for completeness. There's an immediate problem where you state "Suppose $C\neq B$. Then $\exists x \in C$ such taht $x\in C\setminus B$..." This is not strictly true, for example if $C\subseteq B$. It would be more correct to say "Without loss of generality, suppose $\exists x\in C$ such that $x\in C\setminus B$..." If you're not sure what is meant by the phrase "Without loss of generality," it's basically a shorthand way to say "There are multiple cases, but they all follow exactly the same argument so we may as well only consider one." In this instance, the two cases are $\exists x \in C\setminus B$ and $\exists x \in B\setminus C$. These two cases are not mutually exclusive but they are exhaustive and there's functionally no difference in choosing one over the other.
Second, you state "Consequently, this means that $x\in A\setminus B$ or $x\in B\setminus A$." It's important to note that this is only a consequence of "if $x\in A\Delta C$..." which at this point in the proof you haven't shown, as you've made no statements of any kind about whether $x\in A$ or not. As such the conclusion of that paragraph is if $x\in C\setminus B$, then $x\in A$, and not merely $x\in A$ as you seem to have it.
The same error propagates through to the next paragraph, where you seem to again take for granted $x\in A\Delta C$ and detach the conclusion. Moreover, because you have an implicit unrecognized assumption, your contradiction of $x\in A$ and $x\notin A$ doesn't actually show $B=C$, all it shows is $x\notin A\Delta C$.
To rectify the situation, you have to be especially careful in proofs by contradiction about what are assumptions being introduced and what are deductions drawn from previous assumptions and deductions.
I'm not entirely sure this proof can be rectified in this manner though I could be missing something. I think the easiest path forward would be to break into subcases based on $x\in A$ and $x\notin A$ and see where that takes you.