Prove that $\overline{A\setminus B} \cap \left(\overline{A}\cup\overline{B}\right)=\overline{A}$ 
Prove that
$$\overline{A\setminus  B} \cap \left(\overline{A}\cup\overline{B}\right)=\overline{A}$$

My attempts:
Let $x\in \overline{A\setminus  B} \cap \left(\overline{A}\cup\overline{B}\right)$.
Then
$x\in \overline{A\setminus  B}$ and $x\in\overline{A}\cup\overline{B}$.
Then
$x\in \overline{A}\cup B$ and $x\in\overline{A}\cup\overline{B}$.
Then $x\in\overline{A}\cup B\cup\overline{B} $
Then $x\in\overline{A}$.
Hence, $$\overline{A\setminus  B} \cap \left(\overline{A}\cup\overline{B}\right)\subset \overline{A}$$
Is it correct?
I need help to prove that
$$\overline{A}\subset \overline{A\setminus  B} \cap \left(\overline{A}\cup\overline{B}\right)$$
 A: What you have so far looks mostly good -- perhaps elaborate a bit on why $x \in \overline{A \setminus B}$ implies $x \in \overline A \cup {B}$. Also, when you said $x \in A \cup B \cup \overline B$, I think you meant $x \in A \cup (B \cap \overline B)$ -- look over that step.
The other direction starts off as follows. First assume $ x\in\overline A$. Then you want to show it's in the intersection of two things -- $\overline{A \setminus B}$ and $\overline A \cup \overline B$ -- so you should separately show it's in each of those.
A: Let $x \in \overline{A} \implies x \in \overline{A} \cup \overline{B}$. Also,
$x \in \overline{A} \cup B = \overline{A} \cup \overline{B^c} = \overline{A \cap B^c} = \overline{A \setminus B} \implies x \in (\overline{A} \cup \overline{B}) \cap \overline{A \setminus B}$.
A: You can do this completely algebraically:
$$\overline{A \setminus B} \cap (\overline{A} \cup \overline{B})=
\overline{A \cap \overline{B}} \cap (\overline{A} \cup \overline{B})=
(\overline{A} \cup B) \cap (\overline{A} \cup \overline{B})= \overline{A}\cup(B \cap \overline{B})=\overline{A} \cup \emptyset = \overline{A}$$
A: Take the complement of both sides and use DeMorgans rules.
