If $|A\cap B|= \varnothing $ (disjoint sets),
then $|A \cup B|=|A|+|B|$
Using this result alone, prove $|A\cup B| = |A| + |B| - |A\cap B|$
$|A\cup B| = |A| + |B - A|$
$|A\cap B| + |B - A| = |B|$,
summing gives
$|A\cup B| = |A| + |B| - |A\cap B|$.
In the same way, prove:
$$|A∪B∪C| = |A| + |B| + |C| - |A∩B| - |B∩C| - |A∩C| + |A∩B∩C|$$
$|A∪B∪C| = |A|+|B-A|+|C-B-A|$
$|A∩B| + |B-A| = |B|$
$|C-A| + |A∩C| = |C|$
$|C-B|+|+|B∩C|=|C|$
$|C-B-A| + |C| = |C-A|+|C-B|+ |A∩B∩C|$
summing the previous five equations gives
$|A∪B∪C| = |A| + |B| + |C| - |A∩B| - |B∩C| - |A∩C| + |A∩B∩C|$