Prove $\overline{B} - (A-\overline{B}) \subseteq \overline{B}$ Prove $\overline{B} - (A-\overline{B}) \subseteq \overline{B}$
Attempt: 
Let $x \in \overline{B} -(A-\overline{B})$, then 
$x \not\in \overline{B} \land x \not\in (A-\overline{B})$.
Then by 
By DeMorgan's Law, we have, $x \in \overline{B}\land  x\not\in(A\cap B)$
By DeMorgan's Law $x \in \overline{B} \wedge x \in (\overline{A} \cup \overline{B})$
By Distributive Law : $(x\in \overline{B}\land x \in \overline{A})\lor (x\in \overline{B}\land x \in \overline{B})$
AND.... I've accomplished nothing. A???
 A: As Gregor de Cillia pointed out in a comment, you really want to make use of the fact that
$$
C-D\subseteq C.
$$
In regards to your problem of proving $\overline{B}-(A-\overline{B})\subseteq \overline{B}$ by means of an element-chasing proof, notice what you got at the first step in your attempted proof (you actually proved it without knowing it): 
$$
x\in \overline{B}-(A-\overline{B}) = x\in\overline{B}\land x\not\in(A-\overline{B}).
$$
You can stop here. You noted that $x\in\overline{B}$ and... (it doesn't matter what follows). Since you just noticed that $x\in\overline{B}$, you know that 
$$
x\in \overline{B}-(A-\overline{B}) \Longrightarrow x\in\overline{B}
$$
or, more compactly,
$$
\overline{B}-(A-\overline{B})\subseteq \overline{B},
$$
which is exactly what you wanted to show.
A: You were mostly there. Just take one more step, since $\;(b\vee a)\wedge b = b\;$.
Tidying things up a little, you have:
$$\begin{align}\bar B \setminus (A \setminus \bar B)
 &= \bar B \cap \overline{(A \setminus \bar B)} & \text{definition of set difference}
\\ &= \bar B \cap \overline{(A \cap B)} & \text{definition of set difference}
\\ &= \bar B \cap (\bar A \cup \bar B) & \text{DeMorgan's Law}
\\ &= \bar B & \text{absorption}
\\[2ex] 
 \therefore
\;&\;\bar B\setminus (A\setminus \bar B)\subseteq \bar B & \text{since } \bar B\subseteq \bar B
\end{align}$$
A: Actually the OP solved it. The last line says $x\in\overline{B}\wedge x\in\overline{A}$ (in which case $x\in\overline{B}$) OR $x\in\overline{B}\wedge x\in\overline{B}$ (in which case $x\in\overline{B}$).
In either case $x\in\overline{B}$, QED.
