Proving that $(A\setminus C)\cap(B\setminus C)\cap(A\setminus B)=\emptyset$ For each $A,B,C$ how would I prove that $(A\setminus C)\cap(B\setminus C)\cap(A\setminus B)=\emptyset$ ?
My thoughts are if $x\in (A\setminus C)\cap(B\setminus C)\cap(A\setminus B)$, then $x\in (A\setminus C)$ and $x\in (B\setminus C)$ and $x\in (A\setminus B)$ so $x \in A$ and $x \not \in C$ and $x \in B$ and $x \not \in C$  and $x \in A$  and $x \not\in B$.
How would I continue from here?

It seems like you have picked up on the main issue via avid19's push in the right direction, but I would encourage you to think about associativity of $\cap$, something that would immediately answer the question for you:
\begin{align}
(Α\setminus C)\cap(B\setminus C)\cap(A\setminus B)&= (A\cap C^C)\cap(B\cap C^C)\cap(A\cap B^C)\tag{by defn.}\\[0.5em]
&= A\cap C^C\cap B\cap C^C\cap A\cap B^C\tag{assoc. of $\cap$}\\[0.5em]
&= A\cap A\cap C^C\cap C^C\cap B\cap B^C\tag{assoc. of $\cap$}\\[0.5em]
&= (A\cap A)\cap (C^C\cap C^C)\cap (B\cap B^C)\tag{parens}\\[0.5em]
&= (A\cap C^C)\cap(B\cap B^C)\tag{simplify}\\[0.5em]
&= (A\cap C^C)\cap\varnothing\tag{simplify}\\[0.5em]
&= \varnothing\tag{simplify}.
\end{align}
That was probably what someone had in mind when designing that exercise; that is, the goal was for you to see how to use the associativity of $\cap$ to conclude that you end up with $\text{something}\cap(B\cap B^C)$, where you can immediately see that this reduces to $\varnothing$, the right-hand side. 
