Show that $A \subseteq B \iff A \subseteq B-(B-A)$ Can someone please verify this?

Show that $A \subseteq B \iff A \subseteq B-(B-A)$

$(\Rightarrow)$ Let $x \in A$.
Then, $x \notin B-A$.
Also, $x \in B$.
Therefore, $x \in B-(B-A)$
So, $A \subseteq B-(B-A)$.
$(\Leftarrow)$ Let $x \in B-(B-A)$.
Then, $x \in B$ and $x \notin (B-A)$.
From the latter, we get $x \notin B$ or $x \in A$.
Since $x \in B$, it must be the case that $x \in A$.
So, $A \subseteq B$.
Therefore, $A \subseteq B \iff A \subseteq B-(B-A)$
 A: Your argument is fine. Also you can use de Morgan's laws mindlessly to realize that...
$$B-(B- A) = B \cap (B - A)^c$$
$$= B \cap (B \cap A^c)^c$$
$$= B \cap (B^c \cup A)$$
$$= (B \cap B^c)\cup (B \cap A)$$
$$= B \cap A$$
So your set $B - (B - A)$ is none other than $B \cap A$.
So you're trying to show $A \subset B \iff A \subset A \cap B$. This might be slightly easier.
A: You've done quite well. 
One issue I note is in the $\Leftarrow$ direction.
You should start by assuming  $A\subseteq\left(B-(B-A)\right)$.
Then let $x \in A$ and we know also $x\in A \implies x \in (B-(B-A))$. So we now have $x \in (B-(B-A))$.
Your proof of that direction starts by letting $x \in B - (B-A)$, when in fact, we should first let $x \in A$. And then of course, of the subset relation, $x \in A \implies x \in (B-(B-A))$, So $x\in B-(B-A)$ follows.
Unpacking this gives us $$\begin{align} x\in A\implies x\in B-(B-A) & \iff x \in B \land (x \notin (B-A)) \\ \\ 
&\iff x \in B\land (x \notin B \lor x \in A)\\ \\ 
&\iff (x\in B \land x\notin B) \lor (x\in B \land x\in A) \\ \\
&\implies  x \in B \land x \in A\\ \\ 
&\therefore \quad x\in A \implies (x\in A \land x \in B)\\ \\
&\therefore \quad x\in A\implies x\in B\\ \\
& \therefore \quad A\subseteq\left(B-(B-A)\right) \implies A\subseteq B\end{align}$$
A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
$Just for comparison, here is an alternative proof where we use the laws of logic to simplify:
$$\calc
A \subseteq B - (B - A)
\calcop{\equiv}{definition of $\;\subseteq\;$; definition of $\;-\;$, twice}
\langle \forall x :: x \in A \;\Rightarrow\; x \in B \land \lnot (x \in B \land x \not\in A) \rangle
\calcop{\equiv}{logic: use $\;x \in A\;$ on right hand side of $\;\Rightarrow\;$; use $\;x \in B\;$ on other side of $\;\land\;$}
\langle \forall x :: x \in A \;\Rightarrow\; x \in B \land \lnot (\text{true} \land \text{false}) \rangle
\calcop{\equiv}{logic: simplify}
\langle \forall x :: x \in A \;\Rightarrow\; x \in B \rangle
\calcop{\equiv}{definition of $\;\subseteq\;$}
A \subseteq B
\endcalc$$
