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.
Several elementary proofs have been given, but I feel like some of them ran away with element chasing, and I'd like to take a different approach.

$(A\cap B)\setminus C$ is a subset of $B$.

$(A\cap C)\setminus B$ is a subset of the complement of $B$.

Hence, $[(A\cap B)\setminus C]\cap[(A\cap C)\setminus B]\subseteq B\cap B^c=\emptyset$

So the left hand intersection is empty, proving the two sets are disjoint.

There's nothing wrong with the element chasing. In fact that's probably pretty good exercise if you're getting used to basic set theory :)