Showing the following inclusion: $\delta (A\cap B)\cap (\bar A \cup (\overline{\delta B})) \subseteq \delta A$ 
The question at hand is to prove the following inclusion: $\delta (A\cap B)\cap (\bar A \cup (\overline{\delta B})) \subseteq \delta A$ for   $A,B \subset \mathbb R^n$

NOTE: For my notation, $\bar X$ is the complement of $X$ and $\delta X$ is the boundary of $X$. 
I can draw a picture of this and make sense of it, but I have issues showing this is true mathematically. 
One of the theorems we know is that $\delta (A \cap B) \subseteq \delta A \cup \delta B$ if that helps.
Attempt at a  solution:
$$\delta (A\cap B)\cap (\bar A \cup (\overline{\delta B})) \subseteq \delta A$$
$$= (\delta (A \cap B) \, \cap \, \bar A) \ \cup \ ( \ \delta(A \cap B) \cap (\overline{\delta B}))$$
$$\subseteq (\delta (A \cap B) \, \cap \, \bar A) \ \cup \ ( \ (\delta A \cup \delta B) \cap (\overline{\delta B}))$$
$$= (\delta (A \cap B) \, \cap \, \bar A) \ \cup \ ( \ (\delta A \cap \overline{\delta B}) \cup (\delta B \cap \overline{\delta B}))$$
$$= (\delta (A \cap B) \, \cap \, \bar A) \ \cup \  (\delta A \cap \overline{\delta B}) $$
But at this point, I feel lost. Perhaps I didn't approach this the right way from the beginning, or I just don't know how to continue. Any help would be greatly appreciated. 
 A: You're actually almost done.  You've shown that $$\delta (A\cap B)\cap (\bar A \cup \overline{\delta B})\subseteq (\delta (A \cap B) \cap \bar A) \cup  (\delta A \cap \overline{\delta B})$$ (though your notation is a bit screwy; you seem to be using $\iff$ in place of equals signs).  To finish, you just have to prove the right-hand side is contained in $\delta A$.  That is, you need to prove that $\delta (A \cap B) \cap \bar A$ and $\delta A \cap \overline{\delta B}$ are each contained in $\delta A$.  These are both not too hard.  If you're stuck on the first one, you may find it useful to remember that $\delta(A)=\operatorname{cl}(A)\cap\operatorname{cl}(\bar{A})$, so you just have to prove that if $x\in\delta (A \cap B) \cap \bar A$ then both $x\in\operatorname{cl}(A)$ and $x\in \operatorname{cl}(\bar{A})$.
A: I'd take some $x\notin\delta A$. It belongs either to the interior or to the exterior of $A$.


*

*Let $x\in\mathop{\mathrm{ext}} A$. Then $x\notin\delta(A\cap B)$ (there's a neighborhood of $x$ without any point from $A$, hence without any point from $A\cap B$.

*Let $x\in\mathop{\mathrm{int}}A$. If $x\notin\delta(A\cap B)$ then it's okay. If $x\in\delta(A\cap B)$ then $x\in\delta B$. Therefore, $x\notin\overline A$ and $x\notin\overline{\delta B}$, which finishes the proof.
