[proofplan]
The proof reduces convergence in probability to a set identity. For each $n$, the two indicator random variables differ exactly at those points $\omega\in\Omega$ that belong to precisely one of $A_n$ and $A$, namely the symmetric difference $A_n\triangle A$. Since the absolute difference of two indicators is always either $0$ or $1$, the event $\{|\mathbb{1}_{A_n}-\mathbb{1}_A|>\varepsilon\}$ equals $A_n\triangle A$ for every $0<\varepsilon<1$ and is empty for every $\varepsilon\ge 1$. This identity is exactly what is needed to translate the definition of convergence in probability into the condition $\mathbb P(A_n\triangle A)\to 0$.
[/proofplan]
[step:Identify the error event with the symmetric difference]
For each $n\in\mathbb N$, define the real-valued random variables
\begin{align*}
X_n:\Omega\to\mathbb R,\quad \omega\mapsto \mathbb{1}_{A_n}(\omega)
\end{align*}
and
\begin{align*}
X:\Omega\to\mathbb R,\quad \omega\mapsto \mathbb{1}_{A}(\omega).
\end{align*}
These maps are measurable because $A_n,A\in\mathcal F$ and $\{0,1\}\subset\mathbb R$ is equipped with the Borel $\sigma$-algebra inherited from $\mathbb R$.
Fix $n\in\mathbb N$ and $\omega\in\Omega$. By the definition of the indicator function, $X_n(\omega)$ and $X(\omega)$ each belong to $\{0,1\}$. Therefore
\begin{align*}
|X_n(\omega)-X(\omega)|=1
\end{align*}
if and only if exactly one of the two statements $\omega\in A_n$ and $\omega\in A$ holds. This condition is precisely $\omega\in A_n\triangle A$. Hence, for every $\varepsilon$ with $0<\varepsilon<1$,
\begin{align*}
\{\,\omega\in\Omega: |X_n(\omega)-X(\omega)|>\varepsilon\,\}=A_n\triangle A.
\end{align*}
For every $\varepsilon\ge 1$, the same absolute difference is never larger than $\varepsilon$, so
\begin{align*}
\{\,\omega\in\Omega: |X_n(\omega)-X(\omega)|>\varepsilon\,\}=\varnothing.
\end{align*}
[guided]
For each $n\in\mathbb N$, define
\begin{align*}
X_n:\Omega\to\mathbb R,\quad \omega\mapsto \mathbb{1}_{A_n}(\omega)
\end{align*}
and define
\begin{align*}
X:\Omega\to\mathbb R,\quad \omega\mapsto \mathbb{1}_{A}(\omega).
\end{align*}
Because $A_n$ and $A$ are events, they belong to $\mathcal F$. The preimage of $\{1\}$ under $X_n$ is $A_n$, and the preimage of $\{1\}$ under $X$ is $A$; similarly, the preimages of $\{0\}$ are complements in $\Omega$. Thus these indicator maps are real-valued random variables.
The key point is that indicators remember only membership. Fix $n\in\mathbb N$ and $\omega\in\Omega$. If $\omega$ belongs to both $A_n$ and $A$, then $X_n(\omega)=X(\omega)=1$, so $|X_n(\omega)-X(\omega)|=0$. If $\omega$ belongs to neither $A_n$ nor $A$, then $X_n(\omega)=X(\omega)=0$, so again $|X_n(\omega)-X(\omega)|=0$. The only remaining cases are the two asymmetric cases: $\omega\in A_n\setminus A$ or $\omega\in A\setminus A_n$. In either case one indicator is $1$ and the other is $0$, so
\begin{align*}
|X_n(\omega)-X(\omega)|=1.
\end{align*}
The set of points where exactly one of $A_n$ and $A$ occurs is, by definition, the symmetric difference
\begin{align*}
A_n\triangle A=(A_n\setminus A)\cup(A\setminus A_n).
\end{align*}
Therefore, for every $0<\varepsilon<1$, the inequality $|X_n(\omega)-X(\omega)|>\varepsilon$ holds exactly when $|X_n(\omega)-X(\omega)|=1$, which holds exactly when $\omega\in A_n\triangle A$. Hence
\begin{align*}
\{\,\omega\in\Omega: |X_n(\omega)-X(\omega)|>\varepsilon\,\}=A_n\triangle A.
\end{align*}
If $\varepsilon\ge 1$, then no value in $\{0,1\}$ is strictly larger than $\varepsilon$, so
\begin{align*}
\{\,\omega\in\Omega: |X_n(\omega)-X(\omega)|>\varepsilon\,\}=\varnothing.
\end{align*}
This is the entire mechanism behind the theorem: convergence in probability measures the probability of an error event, and for indicators that error event is exactly the symmetric difference.
[/guided]
[/step]
[step:Use symmetric difference convergence to prove convergence in probability]
Assume
\begin{align*}
\lim_{n\to\infty}\mathbb P(A_n\triangle A)=0.
\end{align*}
Let $\varepsilon>0$ be arbitrary. If $0<\varepsilon<1$, the identity from the previous step gives
\begin{align*}
\mathbb P(|X_n-X|>\varepsilon)=\mathbb P(A_n\triangle A)\to 0.
\end{align*}
If $\varepsilon\ge 1$, the previous step gives
\begin{align*}
\mathbb P(|X_n-X|>\varepsilon)=\mathbb P(\varnothing)=0
\end{align*}
for every $n\in\mathbb N$. Thus for every $\varepsilon>0$,
\begin{align*}
\lim_{n\to\infty}\mathbb P(|X_n-X|>\varepsilon)=0.
\end{align*}
By the definition of convergence in probability, $X_n\xrightarrow{\mathbb P}X$, that is,
\begin{align*}
\mathbb{1}_{A_n}\xrightarrow{\mathbb P}\mathbb{1}_A.
\end{align*}
[/step]
[step:Use convergence in probability at one threshold to recover symmetric difference convergence]
Conversely, assume
\begin{align*}
\mathbb{1}_{A_n}\xrightarrow{\mathbb P}\mathbb{1}_A.
\end{align*}
Equivalently, with $X_n=\mathbb{1}_{A_n}$ and $X=\mathbb{1}_A$, the definition of convergence in probability gives
\begin{align*}
\lim_{n\to\infty}\mathbb P(|X_n-X|>\varepsilon)=0
\end{align*}
for every $\varepsilon>0$. Choose the fixed threshold $\varepsilon=\frac12$. Since $0<\frac12<1$, the identity from the first step yields
\begin{align*}
\mathbb P(A_n\triangle A)=\mathbb P(|X_n-X|>\tfrac12).
\end{align*}
Taking limits gives
\begin{align*}
\lim_{n\to\infty}\mathbb P(A_n\triangle A)=0.
\end{align*}
This proves the reverse implication and completes the equivalence.
[/step]
