Prove that $V=X\cap(Y\setminus U)$. This is an exercise page 7 from Sutherland's book Introduction to Metric and Topological Spaces.

Suppose that $V,X,Y$ are sets with $V\subseteq X\subseteq Y$ and suppose that $U$ is a subset of $Y$ such that $X\setminus V=X\cap U$.
Prove that $V=X\cap(Y\setminus U)$.

My attempt:
Let $x\in X\cap(Y\setminus U)$. Then $x\in X$ and $x\in Y\setminus U$. So, $x\in X$ and $x\notin U$.
Here the solution given by Sutherland's book argues differently. So I am wondering if I can say: If an element $x$ is in the set $X$ then we can write $x\in V$ and $x\in X\cap V$.
And continuing, we have $x\in V$ and $x\in X\cap V$ and $x\notin U$.
The last two relations can be eliminated. And hence, $x\in V$.
The second part of the proof is to prove conversely that $V\subseteq X\cap(Y\setminus U)$.
I am wondering if the first part of my proof is valid, especially the second sentence.
 A: You start to reason in circles beyond the point "So I am wondering if I can say:..." 
You need to keep in mind what you are trying to do. Namely you want to show that $x\in X\cap (Y\backslash U)\implies x\in V$. So once you've showed that $x\in V$ you can just stop. If I were your teacher I would ask you to elaborate on why $x\in V$ if $x\in X$ and $x\not \in U$. (Can you do that?)  It is true, however you just say it is. I think any teacher would want you to explain why.
A: Using $+$ to mean union, and multiplication to mean intersection,and $-$ for complement (That is $a-b=a\backslash b$) we have $$V=X-(X-V).$$ We also have $$X-V=X U \;\text  { and } X=X Y.$$  Hence $$V=X-(X-V)=X-X U=( X Y) -( X Y)U=X(Y-Y U)=X(Y-U)=$$ $$=X\cap (Y\backslash U).$$ Note that the condition $U\subset Y$ was unnecessary.
A: No, your proof is not valid. The assumption that $x\in X$ does not imply that $x \in V$ and $x \in X \cap V$. For instance, take the sets $V=\{1,2,3\}$ and $X=\{1,2,3,4\}$. Then $4 \in X$, but $4 \notin V$ and $4 \notin V\cap X$. It might help to argue by contradiction for this direction (though it can also be proved directly).
