Is my proof complete/valid? I can't help but think that my proof is missing something. I'm pretty sure I understand how to prove it, I just don't know if I expressed it correctly in words. As I am teaching myself, I would appreciate a second opinion. Is it fine as it is, or is there something I should improve?
Prove that for arbitrarily chosen sets A, B, and C that
$$ (A-C) - (B - C) \subseteq A - B$$ 
My proof:
Let $x \in (A-C) - (B-C)$. That means that $x \in A$ and $x \not\in (B-C)$ which implies $x \not\in B$. Therefore, we can conclude that $(A-C)-(B-C) \subseteq (A-B)$.
 A: As lulu suggests in his(er) comment, it is not correct.
Here's a way you can prove it:
If $x \in (A \setminus C) \setminus (B \setminus C)$, then $x \in A \setminus C$, but $x \notin B \setminus C$; so $x \in A$.
Suppose that $x \in B$.
Then $x \in B \cap C$ because $x \notin B \setminus C$, yielding $x \in C$, in contradiction with $x \in A \setminus C$.
Now you can conclude that $x \in A \setminus B$.
A: Something that might help if you're just starting (even though amrsa already answered): do a step by step substitution. You seem to skip around. E.g. your start is good; Let $x \in (A-C) - (B-C)$. This tell us that $x \in (A-C)$ and $x \not \in (B-C)$ thus $x \in A$ and $x \not \in C$. Notice that I've broken up the pieces. 
What else can I break up? Well, let's look at $x \not \in (B-C)$. First, it's sometimes difficult to think in negatives, so let's remove it (but not forget that it's there!). If $x \in (B-C)$ then $x \in B$ and $x \not \in C$. Now adding back the negative: $\neg ((x\in B) \wedge (x \not \in C))$ now applying the negative: $(x \not \in B) \vee (x \in C)$.
Interesting. What does this tell us? If $x \in (A-C) - (B-C)$ then $x \in A \wedge x \not \in C \wedge (x \not \in B \vee x \in C)$. Well to make this statement true we must have $x \in A$ but $x \not \in C$, which handles the first two requirements. Within the $(x \not \in B \vee x \in C)$ clause we already have that $x \not \in C$, so for $(x \not \in B \vee x \in C)$ to be true we must have $x \not \in B$. 
But that means if $x \in (A-C) - (B-C)$ then $x \in A$ and $x \not \in B$.
Now you can make your conclusion.
A: You can rewrite the set-theory expression as a boolean sentence with simple belong/does not belong terms and work it out by boolean axioms and rules, then translate back to set-theory terms:
$x\in \left((A\setminus C) \setminus(B\setminus C)\right) \iff$
$x\in (A\setminus C) \land x\notin (B\setminus C) \iff$
$\left(x\in A \land x\notin C \right)\land \lnot\left(x\in B\land x\notin C\right) \iff$
$x\in A \land x\notin C \land \left(x\notin B\lor x\in C\right) \iff$
$x\in A \land \left((x\notin C \land x\notin B)\lor (x\notin C \land x\in C)\right) \iff$
$x\in A \land \left((x\notin C \land x\notin B)\lor FALSE\right) \iff$
$x\in A \land (x\notin C \land x\notin B) \iff$
$(x\in A \land x\notin B) \land x\notin C \color{red}\implies$
$x\in A \land x\notin B \iff$
$x\in A\setminus B$
