Is there an easier proof or way to calculate $1[A \cup B]$? I loathe this because
you need a trick to see
$1[A \cup B] = 1 - 1[A \cup B]^C$
$ = 1 - 1[A^C \cap B^C] = 1 - 1[A1 - I[A^C]I[B^C]$
$ = 1 - (1 - I[A])(1 - I[B])$
And is there some intuition behind this?
$P(A \cup B) = P(A) + P(B) - P( A \cap B)$ lead me to try $1[A \cup B] = 1[A] + 1[B] - 1[A \cap B]$. I didn't get anywhere. I don't know if this is true.
Look here for a generalization of the inclusion-exclusion principle:
http://www.compsci.hunter.cuny.edu/~sweiss/resources/inclusion_exclusion.pdf
What you are looking for follows from that.