I can offer 2 proofs:
1) This problem is equivalent to proving:
So we can verify that every element in (A U B U C) gets counted exactly once.
Take any a in A such that a is not in B or C, then a gets counted once in |A| and that's it because it doesn't appear in any intersection. The argument is the same for B and C. So any element that belongs to only one set is counted exactly once.
Take any x in A and B but not in C. This element gets appears once in A, once in B and once in (A n B) (1+1-1=1) so x gets counted exactly once. Same logic applies to (A n C) and (B n C). So any element that appears in exactly one intersection gets counted exactly once.
Take any y in A, B and C. Then this element appears in every term so we have 1+1+1-1-1-1+1=1. So y gets counted only once.
So every element gets counted exactly once, thus the equality holds.
2) The other one is using the rule awkward just said. Good luck.