Q:

Proving that $A\setminus B\subseteq C$ implies $A\setminus C\subseteq B$ .

Theorem. Suppose $A, B$, and $C$ are sets, and $A\setminus B\subseteq C$. Then $A\setminus C\subseteq B$ . 

What I tried 
Proof: 
My try: 
From the statement $A\setminus C\subseteq B$
We infer that $x\in A$ and $x\notin C$ means $x\in B$
Hence by contradiction, we assume that 
$x\notin B$
The statement then becomes 
$x\in A\setminus B$ and $x\notin C$ but it contradicts with $A\setminus B\subseteq C$ since it implies $x\in C$. Is my try correct. Thanks

A:

I honestly don't see a proof of this statement in your answer. You want to prove that $X\subseteq Y$. The simplest way of doing so, would be element chasing. The proof method is as follows,

Let $x\in X$ be some arbitrary element.
Analyze what it means for $x$ to be in $X$.
Make inferences based on that analysis [contradictions and contrapositive parts could be also made here].
Conclude that $x\in Y$.

So in this case, you would have to take some $x\in A\setminus C$, conclude that $x\in A$ but $x\notin C$. Assume towards contradiction that $x\notin B$, therefore $x\in A$ and $x\notin B$, so $x\in A\setminus B$ ... [and I'll let you finish this yourself].

