Here is what I have so far:
Suppose $x \in (A - B) - C$
So $(x \in A \land x \notin B) \land x \notin C$
Then, using the associative law we can say,
$x \in A \land (x \notin B \land x \notin C)$
But here is my problem. I'm not sure what the next step is because $(x \notin B \land x \notin C)$ does not equal $B - A$.
Am I missing something?
So far so good, you have correctly proved that $x \in (A \setminus B) \setminus C \iff x \in A \land (x \not\in B \land x \not\in C)$.
Now you want to prove that this implies that $x \in A \setminus (B \setminus C)$. By definition this is equivalent to $x \in A \land x \not \in B \setminus C$. Again by definition, $x \in B \setminus C \iff x \in B \land x \not\in C$, hence by De Morgan's law, $$x \not\in B \setminus C \iff \neg(x \in B \land x \not\in C) \iff x \not \in B \lor x \in C.$$ Hence you want to prove: $$x \in A \land (x \not\in B \land x \not\in C) \implies x \in A \land (x \not \in B \lor x \in C).$$
But of course you have $x \not\in B \land x \not\in C \implies x \not\in B$, and in turn $x \not\in B \implies x \not \in B \lor x \in C$. So the implication above is true, because if $Q \implies R$, then $P \land Q \implies P \land R$ for any propositions $P$, $Q$ and $R$.
More concretely, $(A \setminus B) \setminus C$ is the set of elements of $A$ that are neither in $B$ nor in $C$. On the other hand, $A \setminus (B \setminus C)$ is the set of elements of $A$, minus the elements in $B$ except those that are in $C$. It's clear that if an element is not in $B$ then it's not in $B \setminus C$, and so the proposition follows. The difference between the two sets is precisely the set of elements that are in $A$ and $C$ but not in $B$; they're in the second set but not the first.