Show that for any sets $A,B$ and $C$ $A\Delta B\subset A\Delta C\cup B\Delta C$. The problem statement is in the title.
I'm proving a problem in class and it's necessary for me to show the above containment. I've drawn some Venn diagrams to make sure the containment actually makes sense and it does, yet I'm having trouble proving this rigorously. 
I know that to begin, we'd let $x\in A\Delta B$ and eventually show that $x\in A\Delta C\cup B\Delta C$. Since $x\in A\Delta B$, $x\in A\setminus B$ or $x\in B\setminus A$. Yet, I don't know how to show $x$ would end up in $C$. 
Also, the proof of this requires proving an "or" statement, that is, showing that $x\in A\Delta C$ or $x\in B\Delta C$. Would supposing $x\notin B\Delta C$ or $x\notin A\Delta C$ help to prove this?
Thanks for any help or feedback!
 A: Cases?
Take $\;x\in A\Delta B=(A\cup B)\setminus (A\cap B)\;$ , and suppose $\;x\in A\;$ (and thus $\;x\notin B\;$, of course).
Case 1: If $\;x\in C\;$ then:
$$(i)\;\;x\in A\cap C\implies x\notin B\cap C\implies x\in B\Delta C$$
$$(ii)\;\;x\notin A\cap C\implies \ldots $$
Continue on.
A: Whenever you want to prove an "or" statement, assume one of the "or"'s is false. In this case, try assuming that $x\notin B\Delta C$, and then prove that $x\in A\Delta C$. 
Here's why this works:
Let $A$ and $B$ be some statements. Consider the statement "(not $A$) implies $B$". This is false exactly when $B$ is false and $A$ is false. In other words, it has the same truth table as "$A$ or $B$". Thus, proving "$A$ or $B$" is the same as proving "(not $A$) implies $B$".
A: Another way to approach, notice $A\Delta B = (A\cup B)-(A\cap B)$
HINT: Let $x\in A\Delta B$
$\implies x\in A\cup B \land x\notin A\cap B$
$\implies x\in A\cup B \land x\in (A\cap B)^C$
$\implies x\in A\cup B\cup C \land x\in (A\cap B)^C \cup C^C$
Remember $x\in A \implies x\in A\cup B$ for whatever set $B$
