Prove $(A \triangle B) \cap (B\triangle C) \cap (C\triangle A) = \emptyset$ This can be proved by assuming that there exists some $x \in (A \triangle B) \cap (B\triangle C) \cap (C\triangle A) $ and then deriving a contradiction by considering each of the cases that arise. 
[$X\triangle Y$ is the symmetric difference of $X$ and $Y$]
Now, this proof ultimately breaks down into six cases which makes it a bit long considering the simple goal. So I was wondering if there was a better way of doing this?
My six case proof goes something like this: 
Suppose to the contrary that $x \in (A \triangle B) \cap (B\triangle C) \cap (C\triangle A) $. Then $x \in (A \triangle B)$ and $ x \in (B \triangle C)$ and $x \in (C\triangle A)$. Since $x \in (C \triangle A)$, $x \in A\backslash C$ or $x \in C\backslash A$.
Case 1: $x \in A\backslash C$. Since $x \in (B \triangle C)$, $x \in B\backslash C$ or $x \in C\backslash B$. Case 1.1: $x\in B\backslash C$... 
Case 2: $x \in C\backslash A$. Since $x \in (B \triangle C)$, $x \in B\backslash C$ or $x \in C\backslash B$. Case 2.1: $x\in B\backslash C$...

Indicator functions are perhaps a good way to go if you want to avoid cases or symmetry arguments. First prove that
$$\chi_{A \triangle B} = (\chi_A+\chi_B \bmod 2)$$
Then show that if $S = (A \triangle B) \cap (B \triangle C) \cap (C \triangle A)$ then we have
$$\begin{align} \chi_S &= (\chi_A+\chi_B)(\chi_B+\chi_C)(\chi_C+\chi_A) \bmod 2 \\ &= \chi_A\chi_B\chi_C + \chi_A^2\chi_B+\chi_A^2\chi_C+\chi_A^2\chi_C\\
& \qquad +\chi_B^2\chi_C+\chi_B^2\chi_A+\chi_B\chi_C^2+\chi_B\chi_C\chi_A \bmod 2 \\
&= \chi_A\chi_B\chi_C + \chi_A\chi_B + \chi_A\chi_C+\chi_A\chi_C \\
& \qquad +\chi_B\chi_C + \chi_B\chi_A + \chi_B \chi_C + \chi_B\chi_C\chi_A \bmod 2\\
&= 2(\chi_A\chi_B\chi_C + \chi_A\chi_B + \chi_A\chi_C + \chi_B\chi_C) \bmod 2 \\
&= 0 \end{align}$$
