$$A\setminus (A\setminus B)=A\cap (\overline{A\cap\overline{B}})=A\cap(\overline{A}\cup\overline{\overline{B}})=$$
$$=A\cap(\overline{A}\cup B)=(A\cap \overline{A})\cup (A\cap B)=\varnothing\cup (A\cap B)=A\cap B$$
This uses De Morgan's Law $\overline{A\stackrel{\cup}\cap B}=\overline{A}\stackrel{\cap}\cup \overline {B}$ and the distributive law $$A\stackrel{\cup}\cap (B\stackrel{\cap}\cup C)=(A\stackrel{\cup}\cap B)\stackrel{\cap}\cup (A\stackrel{\cup}\cap C)$$
Also, $A\setminus B=A\cap\overline{B}$.
This uses the notation $\overline{A}$ for $A^c$.
Another proof:
If $x\in A\setminus(A\setminus B)$, then $$(x\in A)\wedge (x\not\in A\setminus B)\iff (x\in A)\wedge ((x\not\in A)\lor (x\in B))\iff$$
$$\iff (x\in A)\wedge (x\not\in A))\lor ((x\in A)\wedge (x\in B))\iff$$
$$\iff (x\in A)\wedge (x\in B)\iff (x\in A\cap B)$$
You said: "$x\in A$ and $x\not\in A\setminus B$, which means that $x\in A$ and $(x\not\in A \wedge x\in B)$", but it's actually "$(x\not\in A \lor x\in B)$", since $¬(A\cap \overline {B})=\overline{A}\cup B$ by De Morgan's law.
Since you have that $(x\in A)\wedge ((x\not\in A)\lor (x\in B))$, we know that $(x\in A)\wedge (x\in B)$, since $x$ can't be in $A$ and $\overline {A}$ at the same time so that you can ignore the $(x\not\in A)\lor$ in the expression (or you can show like I did in the second proof above).