$\newcommand\de{\delta}$ $\newcommand\ep{\epsilon}$ $\newcommand\al{\alpha}$
Fix any natural $k$. Let $$\de_k:=\inf_{n\ge k}\|f_n 1_{\{|f_n|\le k\}}\|_2$$ and $$\eta_k:=\ep_{k-1}/(4k).$$ We have $$\|u_k\|_2\le2\de_k \tag{1}$$ and $$\eta_k>0.$$ We want to show that there exists a simple function $v_k$ such that $$ \|u_k-v_k\|_2\le\al_k:=\min(\de_k,\eta_k). \tag{2} $$ If $\de_k=0$, then, by (1), $\|u_k\|_2=0$, and hence we may let $v_k:=0$, to have (2).

Otherwise, $\al_k>0$. By the condition $f_n 1_{\{|f_n|\le k\}} \overset{\sigma(L^2,L^2)}{\underset n\longrightarrow} u_k$, we have $|u_k|\le k$ $\mu$-almost everywhere ($\mu$-a.e.). So, we can find a simple function $u_k$ such that $|u_k-v_k|\le\al_k/\sqrt{1+\mu(E)}$ $\mu$-a.e. Then we will have the inequality in (2), as desired.

$\newcommand\de{\delta}$ $\newcommand\ep{\epsilon}$ $\newcommand\al{\alpha}$
Fix any natural $k$. Let $$\de_k:=\inf_{n\ge k}\|f_n 1_{\{|f_n|\le k\}}\|_2$$ and $$\eta_k:=\ep_{k-1}/(4k).$$ We have $$\|u_k\|_2\le2\de_k \tag{1}$$ and $$\eta_k>0.$$ We want to show that there exists a simple function $v_k$ such that $$ \|u_k-v_k\|_2\le\al_k:=\min(\de_k,\eta_k). \tag{2} $$ If $\de_k=0$, then, by (1), $\|u_k\|_2=0$, and hence we may let $v_k:=0$, to have (2).

Otherwise, $\al_k>0$. By the condition $f_n 1_{\{|f_n|\le k\}} \overset{\sigma(L^2,L^2)}{\underset n\longrightarrow} u_k$, we have $|u_k|\le k$ $\mu$-almost everywhere ($\mu$-a.e.). So, we can find a simple function $v_k$ such that $|u_k-v_k|\le\al_k/\sqrt{1+\mu(E)}$ $\mu$-a.e. Then we will have the inequality in (2), as desired.