Show that $ \ B \subset A \iff (A-B) \cup B = A$ Question:
Show that $ \ B \subset A \iff (A-B) \cup B = A$
My attempt:
First we prove $ \ B \subset A  \implies (A-B) \cup B = A$
Assume $ \ B \subset A$
WTS $ \ (A-B) \cup B = A$
First show $ \ (A-B) \cup B \subseteq A$
$ x \in (A-B) \cup B \implies x\in A-B$ or $ \ x \in B$
If $ \ x \in A -B \implies x \in A$. Otherwise $ \ x\in B \implies x\in A$, since $ \ B \subset A$.
Now we show $  A \subseteq \ (A-B) \cup B$
$ x \in A \implies x\in A-B \implies x \in (A-B) \cup B$
Now we prove $ (A-B) \cup B = A  \implies B \subset A $
Assume $ (A-B) \cup B = A $
WTS $ \ B \subset A$
$ x \in B \implies x\in (A-B) \cup B \implies x\in A$, since  $ (A-B) \cup B = A $
I am not quite sure if this approach is correct. Any feedback would be appreciated. 
 A: Some hints:
Preferably, do things globally, I mean not considering elements, but using already proved or obvious inclusions
E.g. for $\Longrightarrow $, you know that $A-B\subset A$, and $B\subset A$ by hypothesis. It is well known that $\;[X\subset Y$ and $X'\subset Y]\;$ imply $X\cup X'\subset Y$. 
Conversely , you only have to prove that, if an element $x\in A$ is not in $B$, it is in ...
For $\Longleftarrow$,  i.e. $(A-B)\cup B=A$ is the hypothesis, just note that we know
for any $X,X'$, we have $X,X'\subset X\cup X'$ by definition.
A: There is one point that is untrue. You mentioned $x \in A \implies x \in A-B$. That is incorrect. 
You may use $A-B = A \cap B^c$ to show that $(A-B)\cup B=A \cup B$. Obviously, $x \in A \implies x \in A \cup B$.
An alternative approach to proving your claim is to prove $B \subseteq A \Leftrightarrow A \cup B =A$. I believe this can easily be proved.
A: A correct way to prove  that $A \subseteq (A$ \ $B ) \cup B$ is this:
Note that $A$ \ $B$  $=A \cap B^c$.
We will use the De Morgan laws and will prove it with contraposition.
Let $x \notin  (A$ \ $B ) \cup B$ then $$x \in (A \cap B^c)^c \cup B^c \Rightarrow x \in (A^c \cup B) \cap B^c \Rightarrow x \in (A^c \cap B^c) \cup (B^c \cap B)=A^c \cap B^c$$ 
Thus $x \notin A \Rightarrow    A \subseteq (A$ \ $B ) \cup B$
A: $\color {blue}{B \subset A  \implies (A-B) \cup B = A}$
$\boxed{(A-B) \cup B\subset A}$
$x\in (A-B) \cup B \implies \left\lbrace \begin{array}l x\in A- B\subset A\\ \text{or}\\x\in B\subset A\end{array}\right.\implies(A-B) \cup B\subset A$
$\boxed{A \subset(A-B) \cup B}$
$x\in A \implies\left \lbrace \begin{array}lx\in A-B\\\text{or}\\x\in A\cap B\implies x\in B\end{array}\right.\implies x\in (A- B)\cup B\implies A \subset(A-B) \cup B$

$\color{blue}{(A-B) \cup B = A  \implies B \subset A}$
$x\in B\implies x\in (A-B) \cup B \implies x\in A\implies B \subset A$
