Show that if $T\in {\cal B}(c_0)$ and $T$ is weakly compact, then $T$ is compact.
My attempt: $T$ is weakly compact, so there is a reflexive space $X$ , and operators $A\in {\cal B}(X,c_0) $ and $B \in {\cal B}(c_0, X)$ such that $T=AB$. By reflexivity of $X$, operators $A,B$ are weakly compact. To show $B$ is compact suppose $\{x_n\}$ is a bounded sequence in $c_0$. There is a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that the sequence $\{Bx_{n_i}\}$ is weakly convergent. suppose $Bx_{n_i}\to y$ (weakly). Using reflexivity of $X$, $Bx_{n_i}\to y$ (norm), which shows $B$ is compact. Furthermore $T$ is compact.
I think my attempt is wrong, because I know on reflexive spaces , every operator is weakly compact, and using my attempt, every weakly compact operator is compact, while ${\cal B}(H)$ is a counterexample for it. I do not know where it were wrong. Please help me. Thanks in advance.