 Solution

A bounded sequence $(x_n)$ lies in a scalar multiple of the closed unit ball of the reflexive Banach space $X$, which is weakly compact. Part (a) gives a subsequence $x_{k_n}\rightharpoonup z$ for some $z\in X$.

The compact operator $T$ is weak-to-weak continuous, so $T{x}_{k_n}\rightharpoonup Tz$. We claim that the convergence is in norm. Otherwise some further subsequence would satisfy $\Vert T{x}_{k_n}-Tz\Vert \ge \epsilon$. Compactness supplies a norm-convergent subsubsequence; its norm limit must also be its weak limit $Tz$, a contradiction. Hence