How to show that $P(A\cup B) = P(A) + P(B) - P(A\cap B)$ 
Say we have a venn diagram with A on the left, intersection C in the middle and B on the right. 
I want to show that $P(A\cup B) = P(A) + P(B)  - P(A\cap B)$
I know what it can be written as A+B+C, but a lot of the proofs  go from:
$P(A\cup B) = P(A) + P(B)  + P(C)$ to
$P(A) + P(B) = P(A\cup B)  + P(C) $
 and then rearrange to  get the proof and get $P(A\cup B) =P(A) + P(B)-  P(C) $ . I just don't get how the left side of the inequality get go from $P(A\cup B)$  to $P(A) + P(B)$. Doesn't that only happen if we assume that they are disjoint. In this case, do we assume that they are disjoint? I know how to write the proof but just having a hard time actually understanding it. 

You seem to be using $A$ to represent both the blue oval and its subset the left hand crescent (similarly $B$ the red oval and the right hand crescent). This will not aid understanding
So let's instead say we have a Venn diagram with crescent $D$ on the left, intersection $E$ in the middle and crescent $F$ on the right where $D,E,F$ are disjoint events and their probabilities can be added


*

*$P(A)=P(D)+P(E)$

*$P(B)=P(F)+P(E)$

*$P(A \cap B)=P(E)$

*$P(A \cup B)=P(D)+P(E)+P(F)$


Now it should be reasonably clear from this final point that $$P(A \cup B)=\big(P(D)+P(E)\big)+\big(P(F)+P(E)\big)-P(E)$$ and so from the previous points that $$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$
