We compute $\mathbb{P}(\theta_0 \in C_n) = \mathbb{P}(W_n(\theta_0) \leq \xi_\alpha)$ and show $W_n(\theta_0) \xrightarrow{d} \chi^2_p$. Decompose \begin{align*} W_n(\theta_0) = \sqrt{n}(\hat{\theta}_n - \theta_0)^\top I(\theta_0) \sqrt{n}(\hat{\theta}_n - \theta_0) + \sqrt{n}(\hat{\theta}_n - \theta_0)^\top (\hat{i}_n - I(\theta_0)) \sqrt{n}(\hat{\theta}_n - \theta_0). \end{align*} For the first term: since $\sqrt{n}(\hat{\theta}_n - \theta_0) \xrightarrow{d} U \sim N(0, I(\theta_0)^{-1})$, letting $V = I(\theta_0)^{1/2}U \sim N(0, I_p)$, the first term converges in distribution to $V^\top V \sim \chi^2_p$. For the second term: $\hat{i}_n - I(\theta_0) \xrightarrow{\mathbb{P}_{\theta_0}} 0$, so by Slutsky's lemma applied to the bounded-in-distribution factor $\sqrt{n}(\hat{\theta}_n - \theta_0)$ and the vanishing factor $\hat{i}_n - I(\theta_0)$, the second term converges to zero in probability. Adding the two terms by Slutsky's lemma gives $W_n(\theta_0) \xrightarrow{d} \chi^2_p$.