Inclusion-Exclusion: For any two events $A$ and $B$, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
I know that $$P(A \cup B) = \frac{|A \cup B|}{\Omega} = \frac{|A \cap B^c| + |B \cap A^c| + |A \cap B|}{\Omega}$$ but I keep getting plus $P(A \cap B)$ instead of the subtraction as the last term.
It is true that $$|A \cup B| = |A \cap B^C| + |A \cap B| + |A^C \cap B| \tag{1}$$ since the disjoint union of the three sets on the right-hand side (RHS) is the set $A \cup B$. Notice also that $$|A| = |A \cap B^C| + |A \cap B| \tag{2}$$ since the disjoint union of the two sets on the RHS of equation $2$ is the set $A$ and that $$|B| = |A \cap B| + |A^C \cap B| \tag{3}$$ since the disjoint union of the two sets on the RHS of equation $3$ is the set $B$. Therefore, if we add and subtract the term $|A \cap B|$ to equation $1$, we obtain \begin{align*} |A \cup B| & = |A \cap B^C| + |A \cap B| + |A^C \cap B| + |A \cap B| - |A \cap B|\\ & = |A| + |B| - |A \cap B| \end{align*} as required.
Hence, \begin{align*} \Pr(A \cup B) & = \frac{|A \cup B|}{|\Omega|}\\ & = \frac{|A| + |B| - |A \cap B|}{|\Omega|}\\ & = \frac{|A|}{|\Omega|} + \frac{|B|}{|\Omega|} - \frac{|A \cap B|}{|\Omega|}\\ & = \Pr(A) + \Pr(B) - \Pr(A \cap B) \end{align*}