The key is to leverage the total variation convergence. Since $\|\Pi_n - \phi_n\|_{\mathrm{TV}} \to 0$ a.s., we have $|\Pi_n(C_n) - \phi_n(C_n)| \leq \|\Pi_n - \phi_n\|_{\mathrm{TV}} \to 0$ a.s. for any fixed sequence of sets $C_n$. Since $\Pi_n(C_n) = 1-\alpha$ by assumption, we get $\phi_n(C_n) \to 1-\alpha$ a.s. Now $\phi_n = \mathcal{N}(\hat{\theta}_n, I(\theta_0)^{-1}/n)$, so $\phi_n(C_n) = \mathcal{N}(\hat{\theta}_n, I(\theta_0)^{-1}/n)(C_n)$. By the asymptotic normality of the MLE, $\sqrt{n}(\hat{\theta}_n - \theta_0) \xrightarrow{d} \mathcal{N}(0, I(\theta_0)^{-1})$ under $P_{\theta_0}$. For the symmetric interval $C_n = \{\nu : |\nu - \hat{\theta}_n| \leq R_n/\sqrt{n}\}$, the coverage is $P_{\theta_0}(\theta_0 \in C_n) = P_{\theta_0}(|\hat{\theta}_n - \theta_0| \leq R_n/\sqrt{n})$. Since $\phi_n(C_n) = P(|Z| \leq R_n I(\theta_0)^{1/2})$ for $Z \sim \mathcal{N}(0,1)$ and $\phi_n(C_n) \to 1-\alpha$, the threshold $R_n I(\theta_0)^{1/2} \to z_{\alpha/2}$ where $z_{\alpha/2}$ is the $(1-\alpha/2)$-quantile of the standard normal. The coverage probability $P_{\theta_0}(\theta_0 \in C_n) = P(|\sqrt{n}(\hat{\theta}_n - \theta_0)| \leq R_n) \to P(|W| \leq z_{\alpha/2}) = 1-\alpha$ for $W \sim \mathcal{N}(0, I(\theta_0)^{-1})$.