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? 

The argument is correct, but the second half could be simplified: since $y\in A$, $y\in C$, you know that $y\in A\cap C$, and since furthermore $y\notin B$, you can immediately conclude that $y\in(A\cap C)\setminus B$.
