I don't understand the argument in the proof of Lemma 6.5 of Kallenberg's Foundations of Modern Probability (2nd edition):
Let $$\mathfrak A:=\{\mathcal F\subseteq\mathcal A:\mathcal F\text{ is a }\sigma\text{-algebra on }\Omega\}.$$ It is clear to me that it is sufficient to show that $$\sup_{\mathcal F\in\mathfrak A}\operatorname E\left[\left|\operatorname E\left[\xi\mid\mathcal F\right]\right|;A\right]\xrightarrow{\operatorname P[A]\:\to\:0}0\tag1.$$
Moreover, if $(\mathcal F_n)_{n\in\mathbb N}\subseteq\mathfrak A$ and $(A_n)_{n\in\mathbb N}\subseteq\mathcal A$ with $\operatorname P[A_n]\xrightarrow{n\to\infty}0$, then it clearly holds $(\operatorname P[A_n\mid\mathcal F_n])_{n\in\mathbb N}\to0$. In particular, there is an increasing $(n_k)_{k\in\mathbb N}\subseteq\mathbb N$ with $$\operatorname P\left[A_{n_k}\mid\mathcal F_{n_k}\right]\xrightarrow{k\to\infty}0\;\;\;\text{almost surely}\tag2$$ and hence $$\operatorname E\left[\left|\operatorname E\left[\xi\mid\mathcal F_{n_k}\right]\right|;A_{n_k}\right]\le\operatorname E\left[|X|\operatorname P\left[A_{n_k}\mid\mathcal F_{n_k}\right]\right]\xrightarrow{k\to\infty}0\tag3.$$
But how do we obtain $(1)$ from that?