Let $\Gamma$ be an infinite set and denote its power set by $\wp(\Gamma)$. A finitely additive measure on $\Gamma$ is a function $\lambda : \wp(\Gamma) \to \mathbb{R}$ such that $\lambda(A \cup B) = \lambda(A) + \lambda(B)$, for all disjoint $A,B \subset \Gamma$.
Let $X$ be a (real) Banach space and denote by $\ell_\infty(\Gamma,X)$ the Banach space of all bounded functions $f : \Gamma \to X$, equipped with the supremum norm. If $X= \mathbb{R}$, we'll denote $\ell_\infty(\Gamma,X)$ simply by $\ell_\infty(\Gamma)$.
The dual space $\ell_\infty(\Gamma)^*$ is isometrically isomorphic to the space of all bounded finitely additive measures on $\Gamma$, equipped with the variation norm.
Given $x = (x_i)_{i \in \Gamma} \in \ell_\infty(\Gamma,X)$ and $A \subset \Gamma$, denote by $x \chi_A$ the element of $\ell_\infty(\Gamma,X)$ defined by $x\chi_A(j)= 0$, if $j \notin A$, and $x\chi_A(j) = x_j$, if $j \in A$.
A sequence $(x^*_n)_{n \geq 1}$ in the dual space of $X$ is weak$^*$-null if $x_n(x) \to 0$, for all $x \in X$.
I'm trying to follow the proof of Lemma 2 from Leung-Räbiger's paper. This lemma states:
Lemma: Let $(x_n)_{n \geq 1}$ be a bounded sequence in $\ell_\infty(\Gamma,X)$ and $(x^*_n)_{n \geq 1}$ be a weak$^*$-null sequence in $\ell_\infty(\Gamma,X)^*$. Define finitely additive measures $(\lambda_n)_{n \geq 1}$ on $\Gamma$ by $\lambda_n(A) = x^*_n(x_n\chi_A)$, for all $A \subset \Gamma$. Then $(\lambda_n)_{n \geq 1}$ is relatively weakly compact in $\ell_\infty(\Gamma)^*$.
They begin by arguing by contradiction. I'm having trouble with the following argument:
If $(\lambda_n)_{n \geq 1}$ is not relatively weakly compact, then we may assume that there exist disjoint subsets $(A_n)_{n \geq 1}$ of $\Gamma$ and $\varepsilon > 0$ such that $|\lambda_n(A_n)| > \varepsilon$ for all $n \geq 1$.
Any help or hint regarding this step of the proof will be highly appreciated.