For sets $A,B,C$, $(A\setminus B)\subset (A\setminus C)\cup (C\setminus B)$

First of all, I am sorry for my bad english, I am from Brazil :-) I have problem with proof for some set theory task.

Here it is:

$A,B,C$ are three sets. Show that:

$$(A\setminus B) \subset (A\setminus C) \cup (C\setminus B)$$

It is clear by looking at diagrams but I do not know how I show! It is not too hard I think...

Thank you for the help,

Igor

Suppose $\;x\in A\setminus B\;$ . We have only two possibilites:
\begin{align*}(1)&\;\;x\in C : \;\;\implies x\in C\setminus B\implies x\in (A\setminus C)\cup (C\setminus B)\\ (2)&\;\;x\notin C :\;\;\implies x\in A\setminus C\implies x\in (A\setminus C)\cup (C\setminus B)\end{align*}
Hint: Take any $x\in A\setminus B,$ which means that $x\in A$ and $x\notin B$. Now consider two cases (and note that there are no other cases to consider):
• $x\in C$
• $x\notin C$
Show that, in either case, $x\in (A\setminus C)\cup(C\setminus B).$
If you take out from $A$ things that are in $B$, what is left is certainly things in $A$ not in $C$, except for things in $C$ that were not in $B$, so if you add the latter you are all set. The formula is nothing more than this.