Convergence of a confidence interval for the variance of a not normal distribution Given $X_{1},\dots,X_{n}$ i.i.d. $\sim N(μ,σ^{2})$, it's known that a confidence interval for the variance with $1-\alpha$ confidence is as it follows $$\sigma^2\in\Biggl(\frac{(n-1)S_{n}^{2}}{\chi^{2}_{n-1,\alpha/2}},\frac{(n-1)S_{n}^{2}}{\chi^{2}_{n-1,1-\alpha/2}}\Bigg)=(A_{n},B_{n}). $$
I'm asked to compute $$\lim_{n \to \infty}P\Bigl(\sigma^2\in(A_{n},B_{n})\Bigr),$$ taking into account that $X_{1},\dots,X_{n},\dots,$ are i.i.d. (but not necessarily normal) and $\mathbb{E}(X_{1}^{4})<\infty.$.
I know that $\sqrt{n} (S_n^2 - \sigma^2) \rightarrow_{d} \text{N}(0, \sigma^4 (\kappa - 1)),$ where $\kappa=\mu_4/\sigma^{4}$ and $\mu_4 = E(X_i -\mu)^4$, but I have no clue in how to continue. Could you give me some ideas?

Denote $\chi^2_{n - 1, \alpha/2}$ and $\chi^2_{n - 1, 1 - \alpha/2}$ by $\xi_n$ and $\eta_n$ respectively.  In the following we show that as $n \to \infty$,
\begin{align}
P[A_n \geq \sigma^2] = P[(n - 1)S_n^2/\sigma^2 \geq \xi_n] \to \Phi\left(-\sqrt{2}\sigma^2z_{\alpha/2}/\tau\right), \tag{1}
\end{align}
where $\Phi$ is the cdf of the standard normal distribution, $z_{\alpha/2} = \Phi^{-1}(1 - \alpha/2), \tau^2 = \sigma^4(\kappa - 1)$.
To prove $(1)$, we first show that
\begin{align}
\xi_n = (n - 1) + \sqrt{2(n - 1)}(z_{\alpha/2} + o(1)). \tag{2} 
\end{align}
To show $(2)$, note that provided $Y_n \sim \chi^2_{n - 1}$, CLT implies that
\begin{align}
Z_n := \frac{Y_n - (n - 1)}{\sqrt{2(n - 1)}}\to_d N(0, 1), 
\end{align}
whence
\begin{align}
\xi_n &= F_{Y_n}^{-1}(1 - \alpha/2) = (n - 1) + \sqrt{2(n - 1)}F_{Z_n}^{-1}(1 - \alpha/2) \\
&= (n - 1) + \sqrt{2(n - 1)}(z_{\alpha/2} + o(1)),  
\end{align}
i.e., $(2)$ holds.
By $\Delta_n := \sqrt{n}(S_n^2 - \sigma^2)/\tau \to_d N(0, 1)$ and Polya's Theorem, we have
\begin{align}
\sup_{x \in \mathbb{R}}|F_{\Delta_n}(x) - \Phi(x)| \to 0 \tag{3}
\end{align}
as $n \to \infty$. It then follows that
\begin{align}
& \left|P[(n - 1)S_n^2/\sigma^2 \geq \xi_n] - \Phi\left(-\sqrt{2}\sigma^2z_{\alpha/2}/\tau\right)\right| \\
=& \left|P[\Delta_n \geq \sqrt{n}\tau^{-1}((n - 1)^{-1}\xi_n - 1)\sigma^2] - \Phi\left(-\sqrt{2}\sigma^2z_{\alpha/2}/\tau\right)\right| \\
\leq & |F_{\Delta_n}(\sqrt{n}\tau^{-1}((n - 1)^{-1}\xi_n - 1)\sigma^2) - 
\Phi(\sqrt{n}\tau^{-1}((n - 1)^{-1}\xi_n - 1)\sigma^2)| + o(1) \\
&+ |\Phi(-\sqrt{n}\tau^{-1}((n - 1)^{-1}\xi_n - 1)\sigma^2) - 
\Phi\left(-\sqrt{2}\sigma^2z_{\alpha/2}/\tau\right)| \\
\leq & \sup_{x \in \mathbb{R}}|F_{\Delta_n}(x) - \Phi(x)| + o(1) \\
&+ |\Phi(-\sqrt{n}\tau^{-1}((n - 1)^{-1}\xi_n - 1)\sigma^2) - 
\Phi\left(-\sqrt{2}\sigma^2z_{\alpha/2}/\tau\right)| \\
\to & 0
\end{align}
as $n \to \infty$. The "$o(1)$" term stands for $P[\Delta_n = \sqrt{n}\tau^{-1}((n - 1)^{-1}\xi_n - 1)\sigma^2)]$, which tends to $0$ as $n \to \infty$. The last step is a consequence of $(2)$ and $(3)$.
By the similar argument, it can be shown that
\begin{align}
P[B_n \leq \sigma^2] = P[(n - 1)S_n^2/\sigma^2 \leq \eta_n] \to \Phi\left(-\sqrt{2}\sigma^2z_{\alpha/2}/\tau\right).
\end{align}
Therefore,
\begin{align}
P[A_n < \sigma^2 < B_n] = 1 - P[A_n \geq \sigma^2] - P[B_n \leq \sigma^2] 
\to 1 - 2\Phi\left(-\sqrt{2}\sigma^2z_{\alpha/2}/\tau\right).  
\end{align}
The above asymptotic result may be verified by considering $X_1, \ldots, X_n \text{ i.i.d. } \sim N(\mu, \sigma^2)$, for which case $\tau^2 = 2\sigma^4$, whence $P[A_n < \sigma^2 < B_n] \to 1 - \alpha$.  On the other hand, it is well-known that $(A_n, B_n)$ is the exact $1 - \alpha$ confidence interval for $\sigma^2$ under the normality condition.
