Proof that $(A-B)\cap C = (A\cap C)-B$. 
Let $A,B,$ and $C$. Prove that $(A-B)\cap C = (A\cap C)-B$.

Suppose $x\in (A-B)\cap C$, then $x$ is in $(A-B)$ and $C$. If $x\in (A-B)$ then $x\notin B$ but $x\in A$ hence $x\in A\cap C$, but not in $B$ which means $x\in (A\cap C)-B$ therefore $(A-B)\cap C \subset (A\cap C)-B$. Now suppose $y\in (A\cap C)-B$ then $y$ is in $A$ and $C$, but not in $B$ which is the same as $y\in (A-B)\cap (C-B)$, but $ (A-B)\cap (C-B) \subset (A-B)\cap C$ therefore $(A\cap C)-B \subset (A-B)\cap C$.
Did I do this correctly? 

Your argument is correct, but it could be simplified using
$$X - Y = X \cap Y^c, \tag{$\spadesuit$}$$
where $Y^c$ is the complement of $Y$. I urge you to remember $(\spadesuit)$, it is usually very helpful with all kinds of problems with set difference. Then your formula 
$$(A-B) \cap C = (A \cap C) - B$$
becomes
$$ (A\cap B^c) \cap C = (A \cap C) \cap B^c, $$
which is trivially true, because $\cap$ is commutative $X \cap Y = Y \cap X$
and associative $$(X \cap Y) \cap Z = X \cap (Y \cap Z).$$
Using this we conclude with
$$ (A\cap B^c) \cap C = A\cap (B^c \cap C) = A \cap (C \cap B^c) = (A \cap C) \cap B^c. $$
I hope this helps $\ddot\smile$
