Let $C$ and $D$ be subsets of $X$. Prove that $$(X\setminus C)\cap D=D\setminus C$$ My workings: Showing two sets are equal amounts to showing inclusion both ways so I want to show $$(X\setminus C)\cap D\subseteq C$$ and $$(X\setminus C)\cap D\supseteq D\setminus C$$ For the first one I did the following. Let $x\in (X\setminus C)\cap D$, then $x\in(X\setminus C)$ and $x\in D$. This implies that $x\in X$, $x\notin C$ and $x\in D$. Because $D\subseteq X$ this is equivalent to saying $x\in D$, $x\notin C$. Therefore $x\in D\setminus C$. this shows that $(X\setminus C)\cap D\subseteq D\setminus C$.
For the second part I ran into some trouble. This time let $x\in D\setminus C$. Then I thought since $D\subseteq X$ this implies $D\cap X=D$ and we can say $x\in (X\cap D)\setminus C$. I just don't know if this is completely correct to say, and it also doesn't finish the argument.
If anyone could help me out with this last part I would be grateful. Also it is very possible that the arguments in the first are not rigorous enough, if so please feel free to point out where I could improve. Thanks