Notice, $A=B ⇔ ((\forall x \in A ⇒x\in B) \land (\forall x \in B⇒x\in A))$. Looking at the definition you can notice that the right hand side is nothing but two two directions of subset.
Therefore, suppose $A, B$ are two sets, then to show $A=B$ it suffices to show $(A \subseteq B) \land (B \subseteq A)$.
Now, In order to show $A\subseteq B$. Assume an arbitrary $x \in A$, then show $x\in B$.
I will show one direction of subset for the first one, the other is quite similar I hope you will be able to do it.
Suppose $x\in ((X\cup Y)\setminus(Y\cap Z))$. Then, $(x\in X \lor x\in Y) \land (x\notin Y\land x\notin Z)$
As you can see, the above atomics follow from the definition of union, intersection, and difference.
Since $(x\in X \lor x\in Y) \land (x\notin Y\land x\notin Z)$, then $((x\in X\land x\notin Y)\lor (x\in Y\land x\notin Z))$. I got these by De Morgan distribution.
Finally, Translate the definitions into the respective symbol:$((x\in X\land x\notin Y)\lor (x\in Y\land x\notin Z))≡x\in ((X\setminus Y)\lor (Y\setminus Z))$
Which implies: $x\in ((X\setminus Y)\cup(Y\setminus Z))$.
Therefore, $((X\cup Y)\setminus (Y\cap Z))\subseteq ((X\setminus Y)\cup(Y\setminus Z))$
Here is another proof with additional explanations
Show that $(X\cup Y)\setminus (Y\cap Z)=(X\setminus Y)\cup(Y\setminus Z)$
Suppose: $$x\in (X\cup Y)\land x\notin (Y\cup Z)$$
Above supposition follows from the definition of difference.$$x\in(X\cup Y)\rightarrow (x\in X \lor x\in Y)\land x\notin (Y\cap Z)\rightarrow (x\notin Y \lor x\notin Z)$$$$Supp. x\in X \land x\notin Y\rightarrow x\in(X\setminus Y)(a)$$$$Supp.x\in X\land x\notin Z\rightarrow x\in(X\setminus Z)(b)$$$$Supp.x\in Y\land x\notin Y\rightarrow \bot$$$$Supp.x\in Y \land x\notin Z \rightarrow x\in (Y\setminus Z)(c)$$
Therefore, from a and c, $$x\in (X\setminus Y)\cup (Y\cap Z)\rightarrow (X\cup Y)\setminus (Y\cap Z)\subseteq(X\setminus Y)\cup(Y\setminus Z)$$
The other direction of subset is very similar to this one, hopefully you can do it.
The union and intersection operators are monotone. If $A\subseteq A'$
and $B\subseteq B'$ then $A\cup B\subseteq A'\cup B'$ and $A\cap B\subseteq A'\cap B'$. If $f(A,B)$ is a formal combination of $A$s and $B$s with unions and intersections, then it is monotone: $f(A,B)\subseteq f(A',B')$ under the above
assumptions. But $A\setminus B$ is not a monotone operation.
Best Answer
This is the illustration of $X\Delta Y$:
This is the illustration of $(X\Delta Y)\Delta Z$:
Thus, convince yourself that $X\cup Y\cup Z \setminus ((X\Delta Y)\Delta Z)$ is the set of all elements that belong to exactly two of $X,Y,Z$.