Using disjunction to prove that $A \setminus (A \setminus B) = A \cap B$ The problem is as follows:

Suppose $A$ and $B$ are sets. Prove that $A \backslash (A \backslash B) = A \cap B$.

I've rewritten the problem as a biconditional where $A \backslash (A \backslash B) \longleftrightarrow A \cap B$. Proving the first part, $A \backslash (A \backslash B)  \rightarrow A\cap B$, was easy enough. However, I'm finding the second part, $A\cap B \rightarrow A \backslash (A \backslash B)$, much harder. To prove that $x \in A \land (x \notin A \lor \ x \in B)$, all I have is that $x \in A$ and $x \in B$. Even more puzzling is that I'm suppose to be using a disjunction somewhere.    
I should be clear that I'm not simply trying to show an equivalence but write a logical proof. 
 A: \begin{align}
A\setminus (A\setminus B) &= A\cap (A\setminus B)^C\tag{definition}\\[0.5em]
&= A\cap(A\cap B^C)^C\tag{definition}\\[0.5em]
&= A\cap (A^C\color{red}{\cup} B)\tag{DeMorgan / $\color{red}{\text{disjunction}}$}\\[0.5em]
&= (A\cap A^C)\color{red}{\cup} (A\cap B)\tag{distrib. / $\color{red}{\text{disjunction}}$}\\[0.5em]
&= A\cap B\tag{$A\cap A^C=\varnothing$}
\end{align}


Even more puzzling is that I'm suppose to be using a disjunction somewhere. 

I have just added in $\color{red}{\text{red}}$ where disjunction, i.e. $\cup$, is used to make it clearer. There's really no need to use an element-chasing proof here. Basic set algebra, as above, significantly reduces the time/work we have to put into proving the desired identity. 
A: Proposition

$x \in A\backslash(A\backslash B) \leftrightarrow x \in A∩B$

Definitions needed
$A\backslash B = x \in A \wedge x \notin B$
$A\cap B = x \in A \wedge x \in B$

The Proof:
$$\begin{align} x \in A\backslash(A\backslash B) \leftrightarrow x \in A∩B & \equiv x \in A \wedge \neg (x \in A\backslash B) \leftrightarrow x \in A \wedge x \in B \\  
&\equiv x \in A \wedge \neg (x \in A \wedge \neg (x \in B)) \leftrightarrow x \in A \wedge x \in B     \\
&\equiv x \in A \wedge (x \notin A \vee x \in B) \leftrightarrow x \in A \wedge x \in B\\ \\
&\equiv (x \in A \wedge x \notin A) \vee (x \in A \wedge x \in B) \leftrightarrow x \in A \wedge x \in B \\ \\
&\equiv \bot \vee (x \in A \wedge x \in B) \leftrightarrow x \in A \wedge x \in B \\ \\
&\equiv x \in A \wedge x \in B \leftrightarrow x \in A \wedge x \in B \\ \\
\end{align}$$
A: I'll prove what you have  $x\in A\cap B\rightarrow x\in A\setminus(A\setminus B)$ with its contrapositive, i.e. $x\notin A\setminus(A\setminus B)\rightarrow x\notin A\cap B$, and I hope that this will help you see how to do it 'directly'. The trick is to disjunct your statement with a contradiction (just like in the tautology step below). Alternatively you could also try working with the expression that you want to prove ($x\in A\setminus B$ -try to find what that is equivalent to-) and find out which contradiction you need to disjunct with your original hypothesis ($x\in A\cap B$).
Proof by contrapositive: Suppose $x\notin A\setminus(A\setminus B)$, hence by definition of "$\setminus$",
$\neg(x\in A \wedge x\notin(A\setminus B))$, with De Morgan (and definition of "$\setminus$"),
$x\notin A \vee \neg\neg(x\in A \wedge x\notin B)$, from here, use double negation,
$x\notin A \vee (x\in A \wedge x\notin B)$, apply distributive law,
$(x\notin A\vee x\in A)\wedge (x\notin A\vee x\notin B)$. The first part of the conjunction is a tautology and contributes nothing to the conjunction, therefore the last expression is equivalent to the second part.
$x\notin A \vee x\notin B$, with De Morgan again
$\neg(x\in A\wedge x\in B)$, so finally,
$x\notin A\cap B$
A: No need for jargon like biconditional. You are using a slash symbol that means remove those items. So $A\backslash B$ means those elements of $A$  that remain after removing elements of B from it. That is, those elements in $A$ not shared by $B$.When you remove this set from $A$ (as per the notation $A\backslash(A\backslash B)$) we get the elements $A$ shared with $B$, that is $A\cap B$.
