Question: Please could someone validate this proof

Prove that the sets $(A\cap B)$ \ C and $(A\cap C)$ \ B are disjoint. 

First we want to show that 

(1) $(A\cap B)$ \ C $\not \subseteq $ $(A\cap C)$ \ B

Assume $ x \in (A\cap B)$ \ C. Then, by the definition of intersection $ x \in A$ and $x \in B$. We are told that $ x \notin C $. Given that $ x \notin $ C, $ x \notin $ $A\cap C$. Furthermore, we know that $ x \in $ B. So, by the definition of setminus, $ x \notin (A\cap C)$ \ B. This proves (1)

(2) $(A\cap C)$ \ B$\not \subseteq$ $(A\cap B)$\ C 

Assume $ x \in (A\cap C)$ \ B. Then, by the definition of intersection $ x \in A$ and $x \in C$. We know by the definition of setminus that $ x \notin B $. Given that $ x \notin $ B, $ x \notin $ $A\cap B$. Furthermore, we know that $ x \in $ C. So, by the definition of setminus, $ x \notin (A\cap B)$ \ C. This proves (2)

Thanks in advance

Answer: 
You can’t show that $X\cap Y=\varnothing$ by showing that $X\nsubseteq Y$ and $Y\nsubseteq X$: what if $X=\{0,1\}$ and $Y=\{1,2\}$, say?

To show that $(A\cap B)\setminus C$ and $(A\cap C)\setminus B$ are disjoint, just show that if $x\in(A\cap B)\setminus C$, then $x\notin(A\cap C)\setminus B$; that’s all it takes. So suppose that $x\in(A\cap B)\setminus C$; then $x\in A\cap B$, and $x\notin C$. Since $x\notin C$, clearly $x\notin A\cap C$, and it follows at once that $x\notin(A\cap C)\setminus B$.

Note that this is exactly the argument that you gave for your (1). The argument is fine, but it doesn’t show what you thought it showed; instead, it does the whole job for you.
Your logic is sound, but what you stated you are going to prove in (1) and (2) isn't actually what you proved; what you proved is (correctly):

(1) $\forall x(x \in (A \cap B) \setminus C \to x \not\in (A \cap C) \setminus B)$

and 

(2) $\forall x(x \in (A \cap C) \setminus B \to x \not\in (A \cap B) \setminus C)$

Generally, $X$ and $Y$ being disjoint sets is *by definition* the two statements

(1) $\forall x(x \in X \to x \not\in Y)$

and

(2) $\forall x(x \in Y \to x \not\in X)$

so the breakdown into the two sub-proofs is a good idea.