Proving $A \cup (A \setminus B) = A$ I am trying to prove $A \cup (A \setminus B) = A$. Here is my attempt.
\begin{align*}
x \in A \cup (A \setminus B) & \iff x \in A \vee x \in (A \setminus B) \\
& \iff x \in A \vee \left(x \in A \wedge x \not \in B\right) \\
& \iff \left(x \in A \vee x \in A\right) \wedge \left(x \in A \vee x \not \in B\right) \\
& \iff x \in A \wedge \left(x \in A \vee x \not \in B\right) \\
& \iff \left(x \in A \wedge x \in A\right) \vee \left(x \in A \wedge x \not \in B\right) \\
& \iff x \in A \vee \left(x \in A \wedge x \not \in B\right) \\
& \iff x \in A 
\end{align*}
I feel that there are some extraneous steps here. How does this look?
 A: Almost all the steps you write are extraneous. Your line $2$ is identical to line $6$.
On the other hand, you don't really explain how you get from line $6$ to line $7$ (the conclusion), so to be honest, it doesn't look too good.

Instead, what you could do is prove the equivalence in one way, and then the other. In other words, prove that $A\subseteq A\cup (A\setminus B)$, and prove that $A\cup (A\setminus B)\subseteq A$.  At least one of the two relations should be more or less trivial to prove.
Alternatively, you could also

*

*Remember (or, alternatively, prove) that if $C\subseteq A$, then $C\cup A = A$

*Notice that $A\setminus B\subseteq A$
and the conclusion follows directly.
A: As has been pointed out, your rewriting ended up back where you started, and 5xum suggests a probably better way. In case you are interested though, it's also possible, but not easy, to do it your way through algebraic rewriting.
Here is how:
\begin{align*}
&x \in A \vee \left(x \in A \wedge x \not \in B\right) \\
&\iff (x \in A \land (x \in B \lor x \notin B)) \lor (x \in A \land x \notin B) \\
&\iff (x \in A \land x \in B) \lor (x \in A \land x \notin B) \lor (x \in A \land x \notin B) \\
&\iff (x \in A \land x \in B) \lor (x \in A \land x \notin B) \\
&\iff x \in A \land (x \in B \lor x \notin B) \\
&\iff x \in A
\end{align*}
The hard part is going from the first to the second line, where we have to insert a conjunction with $(x \in B \lor x \notin B)$ (applying the law of the excluded middle). The reason to add this is that we somehow need to get rid of the irrelevant $x \notin B$ constraint in the second part of the expression, but it doesn't cancel out with anything currently, so we need to insert something that might cancel with it (some expression involving $B$). Or you can think of it as casework on whether $x$ is in $B$ or not.
A: After the first $\iff$ sign, you have$$x\in A\vee x\in(A\setminus B)\tag1$$and, after the sixth one, you have$$x\in A\vee(x\in A\wedge x\notin B),$$ after which you jump right away to $x\in A$. Why not to go directly from $(1)$ to $x\in A$? After all $x\in A\setminus B\implies x\in A$ and therefore $(1)\implies x\in A$; the reverse implication is trivial.
I would do it as follows: if $A\subset A\cup(A\setminus B)$ and, since $A\setminus B\subset A$, $A\cup(A\setminus B)\subset A$.
A: Your second and second last lines are identical, so everything between is unnecessary.
However, while the next step is valid, it is basically what you were trying to prove.
So this is what you may have been trying for:
$$\begin{align}A\cup(A\smallsetminus B)&\iff A\cup(A\cap B^\complement)\\&\iff (A\cap(B\cup B^\complement))\cup(A\cap B^\complement)\\&\qquad\vdots\\&\qquad\vdots\\&\iff A\cap (B\cup B^\complement)\\&\iff A\end{align}$$
