[proofplan]
We decompose the union $A\cup B$ into the disjoint events $A$ and $B\setminus A$. We also decompose $B$ into the disjoint events $A\cap B$ and $B\setminus A$. Finite additivity of the probability measure on these two disjoint decompositions gives two identities, and eliminating the common term $\mathbb P(B\setminus A)$ yields the addition formula.
[/proofplan]
[step:Decompose the union and one event into matching disjoint pieces]
Define the event $C\in\mathcal F$ by
\begin{align*}
C:=B\setminus A=B\cap A^c.
\end{align*}
Since $A,B\in\mathcal F$ and $\mathcal F$ is a $\sigma$-algebra, we have $A^c\in\mathcal F$, hence $C=B\cap A^c\in\mathcal F$. Also $A\cap B\in\mathcal F$.
The sets $A$ and $C$ are disjoint because
\begin{align*}
A\cap C=A\cap B\cap A^c=\varnothing.
\end{align*}
Moreover,
\begin{align*}
A\cup C=A\cup (B\setminus A)=A\cup B.
\end{align*}
The sets $A\cap B$ and $C$ are disjoint because
\begin{align*}
(A\cap B)\cap C=A\cap B\cap B\cap A^c=\varnothing.
\end{align*}
Finally,
\begin{align*}
(A\cap B)\cup C=(A\cap B)\cup (B\cap A^c)=B.
\end{align*}
[guided]
The purpose of this step is to isolate the part of $B$ that has already been counted inside $A$. Define the event $C\in\mathcal F$ by
\begin{align*}
C:=B\setminus A=B\cap A^c.
\end{align*}
This is an event because $A,B\in\mathcal F$, the $\sigma$-algebra $\mathcal F$ is closed under complements, so $A^c\in\mathcal F$, and it is closed under finite intersections, so $B\cap A^c\in\mathcal F$. The intersection $A\cap B$ is also an event by the same finite-intersection closure.
Now $C$ is exactly the part of $B$ outside $A$. Therefore it has no point in common with $A$:
\begin{align*}
A\cap C=A\cap B\cap A^c=\varnothing.
\end{align*}
It follows that $A\cup B$ is the disjoint union of $A$ and $C$:
\begin{align*}
A\cup C=A\cup (B\setminus A)=A\cup B.
\end{align*}
We also need a decomposition of $B$ using the same remainder $C$. The part of $B$ inside $A$ is $A\cap B$, and the part of $B$ outside $A$ is $C=B\cap A^c$. These two pieces are disjoint because
\begin{align*}
(A\cap B)\cap C=A\cap B\cap B\cap A^c=\varnothing.
\end{align*}
Together they exhaust $B$:
\begin{align*}
(A\cap B)\cup C=(A\cap B)\cup (B\cap A^c)=B.
\end{align*}
[/guided]
[/step]
[step:Apply finite additivity and eliminate the overlap term]
By finite additivity of a probability measure on disjoint events, applied to the disjoint events $A$ and $C$,
\begin{align*}
\mathbb P(A\cup B)=\mathbb P(A\cup C)=\mathbb P(A)+\mathbb P(C).
\end{align*}
Applying finite additivity again to the disjoint events $A\cap B$ and $C$ gives
\begin{align*}
\mathbb P(B)=\mathbb P((A\cap B)\cup C)=\mathbb P(A\cap B)+\mathbb P(C).
\end{align*}
Since all probabilities are real numbers in $[0,1]$, we may subtract $\mathbb P(A\cap B)$ from the second identity to obtain
\begin{align*}
\mathbb P(C)=\mathbb P(B)-\mathbb P(A\cap B).
\end{align*}
Substituting this value of $\mathbb P(C)$ into the first identity yields
\begin{align*}
\mathbb P(A\cup B)
&=\mathbb P(A)+\mathbb P(C)\\
&=\mathbb P(A)+\mathbb P(B)-\mathbb P(A\cap B).
\end{align*}
This is the desired formula.
[guided]
We now use the defining additivity property of the probability measure $\mathbb P$. Finite additivity applies because the sets involved are events in $\mathcal F$ and were shown in the previous step to be disjoint.
First apply finite additivity to $A$ and $C$. Since $A\cap C=\varnothing$ and $A\cup C=A\cup B$, we get
\begin{align*}
\mathbb P(A\cup B)=\mathbb P(A\cup C)=\mathbb P(A)+\mathbb P(C).
\end{align*}
This identity says that measuring $A\cup B$ counts all of $A$ plus only the new part of $B$ outside $A$.
Next apply finite additivity to $A\cap B$ and $C$. Since $(A\cap B)\cap C=\varnothing$ and $(A\cap B)\cup C=B$, we get
\begin{align*}
\mathbb P(B)=\mathbb P((A\cap B)\cup C)=\mathbb P(A\cap B)+\mathbb P(C).
\end{align*}
This second identity expresses the same remainder $C$ in terms of $B$ and the overlap $A\cap B$.
Because probabilities lie in $[0,1]$, these are ordinary real-number equalities. Subtracting $\mathbb P(A\cap B)$ from both sides of the second identity gives
\begin{align*}
\mathbb P(C)=\mathbb P(B)-\mathbb P(A\cap B).
\end{align*}
Substituting this expression for $\mathbb P(C)$ into the identity for $\mathbb P(A\cup B)$ gives
\begin{align*}
\mathbb P(A\cup B)
&=\mathbb P(A)+\mathbb P(C)\\
&=\mathbb P(A)+\mathbb P(B)-\mathbb P(A\cap B).
\end{align*}
Thus the probability of the union equals the sum of the two probabilities with the overlap subtracted once.
[/guided]
[/step]
