How am I to interpret this result? Prove of give counter example: Let A and B be sets.
$$ A \backslash ( A \backslash B) = B \backslash ( B \backslash A) $$
An attempt at a proof:
If $ x \in  A \backslash ( A \backslash B)$ then $x \in A$ and $ x \notin A \backslash B$.
But if $ x \notin A \backslash B$ then either $x \notin A,B$ or $x \in B$ which is a contradiction. 
Am I wrong?  Is there an obvious counter example?
 A: $$A\backslash (A\backslash B)=A\cap (A\cap B^c)^c=A\cap(A^c\cup B)=(A\cap A^c)\cup (A\cap B)=A\cap B,$$
$$B\backslash (B\backslash A)=B\cap (B\cap A^c)^c=B\cap(B^c\cup A)=(B\cap B^c)\cup (B\cap A)=B\cap A= A\cap B,$$
and thus both are equal.
To answer to your question:
There is no contradiction. If $x\in A\backslash (A\backslash B)$, then $x\in A$ but $x\notin A\backslash B$. Therefore $x\in B$ too. Indeed, if $x\notin A\backslash B$, then $x\in A^c$ or $x\in B$. But if $x\in A$, then $x$ can't be in $A^c$, therefore $x$ must be in $B$. We conclude that $x\in A$ and $x\in B$, what we can traduce by $x\in A\cap B$.
A: It's better use algebra to prove it. 
\begin{align}
A-(A-B)&=A\cap(A\cap B^c)^c \hspace{14 mm} (A-B=A\cap B^c)
\\
&=A\cap(A^c\cup B) \hspace{16 mm} ((B^c)^c=B)
\\
&=(A\cap A^c)\cup (A\cap B) \hspace{4 mm} (A\cap (B\cup C)=(A\cap B)\cup(A\cap C))
\\
&=\emptyset\cup(A\cap B) \hspace{19 mm} (A\cap A^c=\emptyset)
\\
&=A\cap B
\end{align}
Similarly there is
$$
B-(B-A)=B\cap A=A\cap B
$$
And so
$$
B-(B-A)=A-(A-B)
$$
A: There is no contradiction.   In point of order, $x\notin A\setminus B$ means either $x\notin A$ or $x\in B$.   That is not an "and".
So if we take any $x$ in $A\setminus(A\setminus B)$, then we have "any $x$ in A except those not also in $B$".  Which is simply "any $x$ only in both $A$ and $B$".
$$A\setminus(A\setminus B) = A\cap B$$
