Lemma about the probability space P (From Grimmett and Stirzaker) This is from Grimmett and Stirzaker, Chapter 1, page 6.

The triple $(\Omega,\mathscr{F},P)$ is a probability space. It has some important properties.
(a) $P(A^C)=1-P(A)$
(b) if $B\supseteq{A}$ then $P(B)=P(A)+P(B\setminus{A})\ge{P(A)}$
(c) $P(A\cup{B})=P(A)+P(B)-P({A}\cap{B})$

The proofs for (b) and (c) are not entirely clear to me.

(b) $B={A}\cup({B}\setminus{A})$. This is the union of disjoint sets and therefore,
$P(B)=P(A)+P({B}\setminus{A})$

I get the above proof. Moving onto the proof for Lemma(c)...

(c) $A\cup{B}=A\cup({B}\setminus{A})$, which is a disjoint union. Therefore by (b),
$P({A}\cup{B})=P(A)+P(B\setminus{A})=P(A)+P({B}\setminus({A}\cap{B}))=P(A)+P(B)-P({A}\cap{B})$

I don't get how (c) is being deduced. Moreover, the start of the proof didn't make sense. Any enlightenment would be appreciated!
 A: $$
A∪(B-A)=A\cup (B\cap A^c)=(A\cup B)\cap (A\cup A^c)=(A\cup B)\cap \Omega=A\cup B\tag1
$$
Also 
$$
A\cap(B-A)=A\cap (B\cap A^c)=(A\cap A^c)\cap B=\varnothing\cap B=\varnothing\tag2
$$
So $A$ and $B-A$ are disjoint. 
Moreover
\begin{align}
B-(A\cap B)&=B\cap(A^c\cup B^c)
\\
&=(B\cap A^c)\cup (B\cap B^c)
\\
&=(B\cap A^c)\cup \varnothing
\\
&=B-A
\end{align}
So
$$
P(A∪B)=P(A)+P(B-A)=P(A)+P(B-(A∩B))\tag3
$$
By replacing $A$ with $B$ and $B-A$ with $A∩B$ in $(1)$, there is
$$
(B-(A∩B))\cup (A∩B)= B\cup (A∩B)=B
$$
$B\cup (A∩B)=B$ for $A∩B\subset B$. Likewise by $(2)$
$$
(B-(A∩B))\cap (A∩B)= \varnothing
$$
Thus
$$
P(B-(A∩B))+P(A∩B)=P(B)
$$
From $(3)$ we have
$$
P(A∪B)=P(A)+P(B)-P(A∩B)
$$
A: First, it is claimed that $$A \cup B = A \cup (B \setminus A)\text{.}$$ This is quite clear from a Venn diagram, but is easy to show formally as well: $x \in A \cup B$, then either $x \in A$, and then it's surely in $A \cup (B \setminus A)$ as well, or $x \notin A$, but then $x \in B$ (to be in $A \cup B$) and so $x \in B \setminus A$, so in $A \cup (B \setminus A)$ as well. Right to left inclusion is clear, as $ A \subseteq A$ and $B \setminus A \subseteq B$, so $A \cup B \subseteq A \cup (B \setminus A)$.
The union is disjoint, because if $x \in A$ it cannot be in $B \setminus A$.
So $P(A \cup B) = P(A \cup (B \setminus A)) = P(A) + P(B \setminus A)$, using the disjointness.
Now, we note that $$B \setminus A = B \setminus (A \cap B)\text{.}$$ This is clear, because if $x \in B \setminus A$, $x \in B$ but not $x \in A \cap B$ and so $x$ is in the right hand side. And if $x \in B \setminus (A \cap B)$, then $x \in B$, and $x \notin A \cap B$. The last can happen only if $x \notin A$, as already $x \in B$. So $x \in B \setminus A$.
As $A \cap B \setminus B$, $P(B) = P(A \cap B) + P(B \setminus (A \cap B))$ by (2), and by the previous equality, $P(B) = P(A \cap B) - P(B \setminus A)$.
The last is rewritten as $P(B \setminus A) = P(B) - P(A \cap B)$.
Now plug this into $P(A \cup B) = P(A) + P(B \setminus A)$ and you're done.
