Inclusion–exclusion for two sets

Partition $A\cup B$ into three pairwise disjoint pieces: $A\setminus B$ (in $A$ only), $B\setminus A$ (in $B$ only), and $A\cap B$ (in both). Since these pieces are disjoint and cover $A\cup B$, $|A\cup B|=|A\setminus B|+|B\setminus A|+|A\cap B|$.

Now observe that $A$ itself splits into the disjoint pieces $A\setminus B$ and $A\cap B$, so $|A|=|A\setminus B|+|A\cap B|$, which means $|A\setminus B|=|A|-|A\cap B|$. Symmetrically, $|B\setminus A|=|B|-|A\cap B|$.

Substituting these two expressions back into the first equation gives $|A\cup B|=(|A|-|A\cap B|)+(|B|-|A\cap B|)+|A\cap B|=|A|+|B|-|A\cap B|$, which is exactly $|A\cup B|=|A|+|B|-|A\cap B|$.