How can I prove this $P(A)=P(A \cap B)+P(A \cap B^c)$? How can I prove this $P(A)=P(A \cap B)+P(A \cap B^c)$ formula?
Is this formula only applicable for disjoint events?
 A: Nothing wrong with Wei's answer, but I started typing this and it is a different approach, so I'm going to post. Note that if you drew a Venn diagram, it would be obvious; just as Wei said, if you have to stick an element e into the set A, it will either be in B or not in B, which is basically all that says.
Anyway, another approach:
As pointed out, I have to split the answer into two cases:
if $P(A) = 0$, then ...
$$P(A \cap B) + P(A \cap B^c) = P(0 \cap B) + P(0 \cap B^c) = 0$$
Otherwise:
$$ P(A \cap B) + P(A \cap B^c) $$
$$ = P(A)P(B \mid A) + P(A)P(B^c \mid A) $$
$$ = P(A)\big( P(B \mid A) + P(B^c \mid A)\big)$$
$$ = P(A)\big( P(B \mid A) + \big(1 - P(B \mid A)\big)\big)$$
$$ = P(A)(1) $$
$$ = P(A) $$
The brackets get confusing, sorry. The main trick here is that the probability of a complement is equal to 1 minus that probability. As Wei says, this is regardless of whether the events are disjoint or not.
A: It applies to any events.
If A and B are joint events, i.e. their intersection is not empty.  Then for any elements e in A, e is either in the intersection or not, thus $P(A)=P(A\wedge B)+P(A\wedge B^c).$ 
If A and B are disjoint, $P(A\wedge B)=0$.  Then $P(A)=P(A\wedge B^c)$.
A: A defining property of a probability measure $P:\mathcal A\to[0,1]$ is:

If $A_1,A_2,\dots$ are disjoint then $P(\bigcup_{n=1}^{\infty}A_n)=\sum_{n=1}^{\infty}P(A_n)$.

Application on $A_n=\varnothing$ for each $n$ leads to $P(\varnothing)=0$.
Then a second application on $A_1=A\cap B$, $A_2=A\cap B^c$ and $A_3=A_4=\cdots=\varnothing$ leads to the formula mentioned in your question.
A: Very simple to answer.Here A intersection B and A intersection B^c are mutually exclusive.Now use the law of addition for mutually exclusive event and then simplify the event.You will get A.Hence,P(A) is our required answer.
A: The Probability third axiom states:
$P\left(\bigcup_{i = 1}^\infty E_i\right) = \sum_{i=1}^\infty P(E_i).$

Set $w.o.g$: 
$E_1 = A \cap B$, $E_2 = A \cap B^c$ and $ E_i=\varnothing \ \ for \ \ i\geq 3$. 

Note that by using set theory one can show that:
$ (A \cap B) \cup (A \cap B^c) = (B \cup B^c) \cap A = A$ see example 1.11.2

Plugging the $E_i$'s into the third axiom and using the claim above:
$P(E_1\bigcup E_2) = P((A \cap B) \bigcup (A \cap B^c)) = P(A) = P(A \cap B) + P(A \cap B^c)$
