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. 

One of the reasons that Venn Diagrams are effective in visualizing probability problems as it poses a rather difficult concept to grasp (probability) in terms of a more intuitive concept (area). So, why don't we use area as a tool to help us prove this concept informally.
Let $P(A)$ be the area enclosed by circle $A$, and $P(B)$ the area enclosed by circle $B$. We seek to find the total area covered by both circles, i.e., $P(A\cup B)$. As you noted, $P(A\cup B)\neq P(A)+P(B)$ since the circles overlap, so we double count the area contained by both circles. How much area is double counted? The area contained by both circles, otherwise denoted by $P(A\cap B)$. This should help you develop intuition for why $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$Now, this is by no means a proper proof, but it should develop your intuition on the issue. And actually, this concept of representing probability as area is not as crazy as it may seem.  As you'll learn if you continue studying probability, probability is closely related to a concept of measure, which is a representation of how much "space" a set contains. 
Another way to see the proof is the following. Denote the area contained by both circles as $P(C)$. So, our equation becomes $$P(A\cup B) = P(A-C)+P(B-C)+P(C)$$Now since $C\subset A$ and $C\subset B$, $$P(A-C)=P(A)-P(C)\quad P(B-C)=P(B)-P(C)$$So, when we substitute into our equation, we get $$P(A\cup B) = P(A)+P(B)-P(C)$$
