Let $\sigma=\langle A_n:n\in\Bbb N\rangle$ be a sequence of subsets of some set $\Omega$. A point $\omega\in\Omega$ is eventually in $\sigma$ if there is an $n_0\in\Bbb N$ such that $\omega\in A_n$ for all $n\ge n_0$, i.e., if $\omega$ is in each member of a ‘tail’ of the sequence. The point $\omega$ is frequently in $\sigma$ if for each $m\in\Bbb N$ there is an $n\ge m$ such that $\omega\in A_n$, i.e., if $\omega$ is in infinitely many members of the sequence. These terms provide an easy way to think and talk about the liminf and limsup of a sequence of sets: $\liminf_nA_n$ is the set of points of $\Omega$ that are eventually in $\sigma$, and $\limsup_nA_n$ is the set of points of $\Omega$ that are frequently in $\sigma$. This is quite easy to verify from the definitions. For example, $\liminf_{n\in\Bbb N}A_n=\bigcup_{n\in\Bbb N}\bigcap_{k\ge n}A_k\;,\tag{1}$ so $\omega\in\liminf_nA_n$ iff there is an $n\in\Bbb N$ such that $\omega\in\bigcap_{k\ge n}A_k$, which is the case iff $\omega\in A_k$ for each $k\ge n$: in short, $\omega\in\liminf_nA_n$ iff $\omega$ is eventually in $\sigma$. Similarly, $\limsup_{n\in\Bbb n}A_n=\bigcap_{n\in\Bbb N}\bigcup_{k\ge n}A_k\;,\tag{2}$ so $\omega\in\limsup_n A_n$ iff for each $n\in\Bbb N$ $\omega\in\bigcup_{k\ge n}A_k$, which is the case iff $\omega\in A_k$ for some $k\ge n$: $\omega\in\limsup_nA_n$ iff $\omega$ is frequently in $\sigma$.
It’s easy to check that $\liminf_{n\in\Bbb N}I_{A_n}(\omega)=\bigvee_{n\in\Bbb N}\bigwedge_{k\ge n}I_{A_k}(\omega)\text{ for all }\omega\in\Omega$ and $\limsup_{n\in\Bbb N}I_{A_n}(\omega)=\bigwedge_{n\in\Bbb N}\bigvee_{k\ge n}I_{A_k}(\omega)\text{ for all }\omega\in\Omega$ are simply restatements of $(1)$ and $(2)$ in terms of indicator functions. (E.g., $\omega$ is eventually in $\sigma$ iff $I_{A_n}(\omega)$ is eventually $1$.) Thus, the following statements are equivalent:
$\begin{align*}&\lim_{n\in\Bbb N}A_n\text{ exists}\tag{3}\\&\liminf_{n\in\Bbb N}A_n=\limsup_{n\in\Bbb N}A_n\tag{4}\\&\liminf_{n\in\Bbb N}I_{A_n}(\omega)=\limsup_{n\in\Bbb N}I_{A_n}(\omega)\text{ for all }\omega\in\Omega\tag{5}\end{align*}$
To finish the proof, you need only show that $(5)$ is equivalent to
$\lim_{n\in\Bbb N}I_{A_n}(\omega)\text{ exists for each }\omega\in\Omega\tag{6}$
and then show that the limit in $(6)$ is the indicator function of the limit in $(3)$.
It’s all just a matter of translating between two ways of saying the same thing: $\omega\in A$ iff $I_A(\omega)=1$, and $\omega\notin A$ iff $I_A(\omega)=0$.