Two-set inclusion–exclusion

The key idea is to split $A \cup B$ into three pairwise-disjoint pieces and count each piece once.

First, partition each set using the other: $A = (A \setminus B) \cup (A \cap B)$ and $B = (B \setminus A) \cup (A \cap B)$. In each equation the two pieces on the right are disjoint, because $(A\setminus B) \cap (A \cap B) = \varnothing$ (an element outside $B$ cannot simultaneously be inside $B$) and likewise $(B\setminus A) \cap (A \cap B) = \varnothing$. Since a finite set's size equals the sum of the sizes of any partition into disjoint pieces, this gives $|A| = |A \setminus B| + |A \cap B|$ and $|B| = |B \setminus A| + |A \cap B|$.

Next, observe that $A \cup B$ itself splits into three pairwise-disjoint pieces: $A \cup B = (A \setminus B) \cup (B \setminus A) \cup (A \cap B)$. Indeed $A\setminus B$, $B\setminus A$, $A\cap B$ are pairwise disjoint (an element in $A\setminus B$ is not in $B$, hence not in $A \cap B$ or in $B \setminus A$; symmetrically for the others), and their union recovers exactly the elements that are in $A$ or in $B$. Counting this partition gives $|A \cup B| = |A \setminus B| + |B \setminus A| + |A \cap B|$.

Finally substitute: from $|A| = |A \setminus B| + |A \cap B|$ we get $|A \setminus B|$ $= |A| - |A\cap B|$, and from $|B| = |B \setminus A| + |A \cap B|$ we get $|B \setminus A|$ $= |B| - |A\cap B|$. Plugging both into $|A \cup B| = |A \setminus B| + |B \setminus A| + |A \cap B|$ gives $|A \cup B| = (|A| - |A\cap B|) + (|B| - |A\cap B|) + |A\cap B| = |A| + |B| - |A\cap B|$, which is exactly $|A \cup B| = |A| + |B| - |A \cap B|$. This finishes the proof.