What I know:
r.v. $X_1, \dots, X_n \sim i.i.d \ Po(\lambda)$. From the central limit theorem, it follows that $$ \sqrt{n}(\bar{X} - \lambda) \xrightarrow{L} \mathcal{N}(0, \lambda). $$ The following follows when a variance stabilization transformation is performed on it. $$ 2\sqrt{n}(\sqrt{\bar{X}}-\sqrt{\lambda})\xrightarrow{L} \mathcal{N}(0, 1). $$ Let $u_{\alpha/2}$ be the upper $α/2$ point of the standard normal distribution.
Let $H_n$ be the CDF of $2\sqrt{n}(\sqrt{\bar{X}}-\sqrt{\lambda})$. Then, from the definition of law convergence, it follows that $$ H_n(u_{\alpha/2}) = P(2\sqrt{n}(\sqrt{\bar{X}}-\sqrt{\lambda}) \leq u_{\alpha/2}) \to \Phi(u_{\alpha/2}) = 1-\frac{\alpha}{2} \\ \therefore P(\sqrt{\bar{X}}-\sqrt{\lambda} \leq \frac{u_{\alpha/2}}{2\sqrt{n}}) \to 1-\frac{\alpha}{2} $$
Question:
Why does this hold true? $$ P(|\sqrt{\bar{X}}-\sqrt{\lambda}| \leq \frac{u_{\alpha/2}}{2\sqrt{n}}) \to 1-\alpha. $$
Taking the complement of $\mathbb P\left(2\sqrt{n}\left(\sqrt{\bar{X}}-\sqrt{\lambda}\right) \leq u_{\alpha/2}\right) \to \Phi(u_{\alpha/2}) = 1-\frac{\alpha}{2}$ you get $\mathbb P\left(2\sqrt{n}\left(\sqrt{\bar{X}}-\sqrt{\lambda}\right) \gt u_{\alpha/2}\right) \to 1-\Phi(u_{\alpha/2}) = \frac{\alpha}{2}$
and similarly for the other tail you get $\mathbb P\left(2\sqrt{n}\left(\sqrt{\bar{X}}-\sqrt{\lambda}\right) \lt -u_{\alpha/2}\right) \to \Phi(-u_{\alpha/2}) = \frac{\alpha}{2}$
and, since they are disjoint events, you can find the probability of their union by summing the individual probabilities $\mathbb P\left(\left|2\sqrt{n}\left(\sqrt{\bar{X}}-\sqrt{\lambda}\right)\right| \gt u_{\alpha/2}\right) \to \alpha$
and for its complement $\mathbb P\left(\left|2\sqrt{n}\left(\sqrt{\bar{X}}-\sqrt{\lambda}\right)\right| \leq u_{\alpha/2}\right) \to 1-\alpha$
which with division by $2\sqrt{n}>0$ gives $\mathbb P\left(\left|\sqrt{\bar{X}}-\sqrt{\lambda}\right| \leq \frac{u_{\alpha/2}}{2\sqrt{n}}\right) \to 1-\alpha$.