Proving : $A \cap (B-C) = (A \cap B) - (A \cap C)$

I have proved this using Venn diagram but when I am trying to prove this using the rule that "If $ A \subset B \text{ and } B \subset A $ then $ A = B $", I am having some problems with my understanding of the same,here is how I did so far:

Let $x \in A \cap (B-C) \Rightarrow x \in A \text{ and } x \in (B-C) \Rightarrow x \in A \text{ and } (x \in B \text{ and } x \notin C) $

How to proceed next? Since if I am do something like this: $ x \in A \text{ and } x \in B \text{ and } x \in A \text{ and } x \notin C $, it's not giving the correct results, what exactly I am missing here?

Hint: $x \notin C \Rightarrow x \notin D \cap C$ for any set $D$.

Here is a different style of proof, viz. an equational and calculational one.

The idea is to start at the most complex side, translate from set theory to logic by working at the element level, then expanding the definitions, and see where there leads you.

In other words, for all $\;x\;$ we calculate: \begin{align} & x \in (A \cap B) - (A \cap C) \\ \equiv & \;\;\;\;\;\text{"expand definition of $\;-\;$, and of $\;\cap\;$ (twice)"} \\ & x \in A \land x \in B \land \lnot(x \in A \land x \in C) \\ \equiv & \;\;\;\;\;\text{"DeMorgan"} \\ & x \in A \land x \in B \land (x \not\in A \lor x \not\in C) \\ \equiv & \;\;\;\;\;\text{"use leftmost conjunct $\;x \in A\;$ in the rightmost conjunct"} \\ & x \in A \land x \in B \land (\textrm{false} \lor x \not\in C) \\ \equiv & \;\;\;\;\;\text{"simplify"} \\ & x \in A \land x \in B \land x \not\in C \\ \equiv & \;\;\;\;\;\text{"reintroduce $\;-\;$ and $\;\cap\;$"} \\ & x \in A \cap (B - C) \\ \end{align}