[proofplan] The assumed convergence in distribution gives the limiting probability that the studentised statistic lies in a symmetric standard normal acceptance region. Since the endpoints $\pm z_{1-\alpha/2}$ are continuity points of the standard normal distribution function, this probability converges to $1-\alpha$. The positivity of the studentising denominator permits an algebraic rearrangement of the event into membership of $a^\top\beta_0$ in the displayed interval. [/proofplan] [step:Convert studentised convergence into a two-sided probability statement] Let $\Phi: \mathbb{R} \to [0,1]$ denote the standard normal distribution function, and let \begin{align*} z := z_{1-\alpha/2} = \Phi^{-1}(1-\alpha/2). \end{align*} Because $\alpha \in (0,1)$, we have $1-\alpha/2 \in (1/2,1)$, so $z \in (0,\infty)$. The standard normal distribution has continuous distribution function $\Phi$, hence both $-z$ and $z$ are continuity points of the limiting distribution. From $T_n \xrightarrow{d} \mathcal{N}(0,1)$, it follows that \begin{align*} \mathbb{P}(-z \leq T_n \leq z) &\to \Phi(z)-\Phi(-z). \end{align*} By the definition of $z$ and the symmetry of the standard normal distribution, \begin{align*} \Phi(z)-\Phi(-z) &= \left(1-\frac{\alpha}{2}\right)-\frac{\alpha}{2} \\ &= 1-\alpha. \end{align*} Therefore \begin{align*} \mathbb{P}(-z \leq T_n \leq z) \to 1-\alpha. \end{align*} [guided] We first translate the convergence in distribution into a probability statement about the event that the statistic falls inside the usual two-sided standard normal acceptance region. Define \begin{align*} z := z_{1-\alpha/2} = \Phi^{-1}(1-\alpha/2), \end{align*} where $\Phi: \mathbb{R} \to [0,1]$ is the standard normal distribution function. Since $\alpha \in (0,1)$, the number $1-\alpha/2$ lies strictly between $1/2$ and $1$, so $z$ is a positive real number. The assumption says that the real-valued random variables $T_n$ converge in distribution to a standard normal random variable. Convergence in distribution gives convergence of probabilities for intervals whose boundary points are continuity points of the limiting distribution function. The standard normal distribution function is continuous everywhere, so the boundary points $-z$ and $z$ are admissible. Hence \begin{align*} \mathbb{P}(-z \leq T_n \leq z) &\to \Phi(z)-\Phi(-z). \end{align*} By the definition of $z$, $\Phi(z)=1-\alpha/2$. By symmetry of the standard normal distribution, $\Phi(-z)=1-\Phi(z)=\alpha/2$. Therefore \begin{align*} \Phi(z)-\Phi(-z) &= \left(1-\frac{\alpha}{2}\right)-\frac{\alpha}{2} \\ &= 1-\alpha. \end{align*} Thus the studentised statistic lies between $-z$ and $z$ with probability tending to $1-\alpha$: \begin{align*} \mathbb{P}(-z \leq T_n \leq z) \to 1-\alpha. \end{align*} [/guided] [/step] [step:Rearrange the two-sided event into interval coverage] For each $n \in \mathbb{N}$, the denominator $S_n(a)$ is positive almost surely by hypothesis. On the event $\{S_n(a)>0\}$, the inequalities \begin{align*} -z \leq \frac{a^\top(\hat{\beta}_n-\beta_0)}{S_n(a)} \leq z \end{align*} are equivalent, after multiplying by $S_n(a)$, to \begin{align*} -zS_n(a) \leq a^\top\hat{\beta}_n-a^\top\beta_0 \leq zS_n(a). \end{align*} Rearranging both inequalities gives \begin{align*} a^\top\hat{\beta}_n-zS_n(a) \leq a^\top\beta_0 \leq a^\top\hat{\beta}_n+zS_n(a). \end{align*} This is precisely the event $\{a^\top\beta_0 \in I_n(a)\}$. Since $\mathbb{P}(S_n(a)>0)=1$, we have \begin{align*} \mathbb{P}\left(a^\top\beta_0 \in I_n(a)\right) = \mathbb{P}(-z \leq T_n \leq z). \end{align*} Combining this equality with the preceding step yields \begin{align*} \mathbb{P}\left(a^\top\beta_0 \in I_n(a)\right) \to 1-\alpha. \end{align*} Therefore $I_n(a)$ is an asymptotic $1-\alpha$ confidence interval for $a^\top\beta_0$. [/step]