(A−B)∪(B−A)=(A∪B) − (A∩B) Please confirm if my following attempt is correct
Let x ∈ (A − B) ∪ (B − A) ⇒ x ∈ (A – B) ∨ x ∈ (B – A)    
⇒ (x ∈ A ∧ x ∉ B) ∨  (x ∈ B ∧ x ∉ A) 
⇒ ((x ∈ A ∧ x ∉ B) ∨ x ∈ B) ∧ ((x ∈ A ∧ x ∉ B) ∨ x ∉ A) 
by distributive law
⇒ ((x ∈ A ∨ x ∈ B) ∧ (x ∉ B ∨ x ∈ B)) ∧
                                 ((x ∈ A ∨ x ∉ A) ∧ (x ∉ B ∨ x ∉ A)) 
                    by distributive law
⇒ ((x ∈ A∪B) ∧ TRUE) ∧ (TRUE ∧ x ∉ (B ∧ A))  (As  P ∨ ¬P = TRUE) 
⇒  x ∈ A∪B) ∧ x ∉ (B ∧ A)      (As P∧TRUE = P)
⇒ x ∈ (A∪B) − (B∩A)
⇒ (A − B) ∪ (B − A) ⊆ (A∪B) − (B∩A) 
Conversely,  Let y ∈ (A∪B) − (A∩B) 
⇒  (y ∈ A ∨ y ∈ B) ∧  y ∉ (A∩B) 
⇒ (y∈A ∨ y∈B) ∧ (y∉A ∨ y∉B)
⇒ ((y∈A ∨ y∈B) ∧ y∉A) ∨ ((y∈A ∨ y∈B) ∧ y∉B)
⇒ ((y∈A ∧ y∉A) ∨ (y∈B ∧ y∉A)) ∨ ((y∈A ∧ y∉B) ∨ (y∈B ∧ y∉B))
⇒ (FALSE ∨ y∈(B−A)) ∨ (y∈(A−B) ∨ FALSE)   (As P ∧ ¬P = FALSE) 
⇒ y∈(B−A) ∨ y∈(A−B)      (As P ∨ FALSE = P)
⇒ y∈((B−A) ∪(A−B)) 
⇒ (A∪B) − (A∩B) ⊆ (A − B) ∪ (B − A)   … (ii)
Therefore from (i) and (ii) we get;
         (A−B)∪(B−A) = (A∪B) − (A∩B)   Proved

 A: This is very complicated for such an easy exercise. Notice that by definition, $$A\setminus B=A\setminus (A\cap B)=A\cap (A\cap B)^c.$$
Therefore, $$(A\setminus B)\cup (B\setminus A)=A\cap (A\cap B)^c\cup B\cap (A\cap B)^c$$
$$\underset{(*)}{=}(A\cup B)\cap (A\cap B)^c=(A\cup B)\setminus (A\cap B),$$
where $(*)$ come from Morgan law.
A: Your proof looks correct, but a much easier way to prove these kind of statements is to use indicator functions. 
Given a set $A$, the indicator function $\mathbf 1_A$ is defined by
$$\mathbf 1_A(x)=\begin{cases}1&\text{if $x\in A$}\\0&\text{if $x\notin A$}.\end{cases}$$
Clearly, $A=B$ if and only if $\mathbf 1_A=\mathbf 1_B$. We can express the indicator functions of union, intersection, etc. in terms of the indicator functions of the individual sets:
\begin{align*}
\mathbf 1_{A\cap B}= \mathbf 1_A\cdot\mathbf 1_B && \mathbf1_{A\cup B}=\mathbf 1_A+\mathbf1_{B}-\mathbf1_A\cdot\mathbf 1_B && \mathbf 1_{A-B}=\mathbf 1_A\cdot(1-\mathbf1_B).
\end{align*}
So for your proof, we do:
\begin{align*}
\mathbf 1_{(A-B)\cup(B-A)} &= \mathbf1_{A-B}+\mathbf 1_{B-A}-\mathbf 1_{A-B}\mathbf 1_{B-A}\\
&=\mathbf 1_A(1-\mathbf 1_B)+\mathbf 1_B(1-\mathbf 1_A)-\mathbf 1_A(1-\mathbf 1_B)\mathbf 1_B(1-\mathbf 1_A)\\
&=\mathbf 1_A+\mathbf 1_B+2\,\mathbf 1_A\mathbf1_B\\
&= \mathbf 1_A + \mathbf 1_B \qquad (\text{$2=0$ in $\mathbb Z_2$}),
\end{align*}
which is the corresponding indicator function for symmetric difference. Indeed, if we simplify $\mathbf 1_{(A\cup B)-(A\cap B)}$ we get $\mathbf 1_A+\mathbf 1_B$, proving that the two are equal.
