Hello Math Stack Exchange. I'm currectly studying the functional analysis, and I am kind of rusty to construct a proof. I have been thinking a lot about the problem, that I have been unsure how to understand in the hint. Here is what the problem says:
Problem A: Let $X$ and $Y$ be infinite dimensional Banach spaces. If $X$ is reflexive and separable, and $T\in \mathcal{L}(X,Y)$ satisfies that $\| Tx_n - Tx\|\to 0$ as $n\to \infty$ whenever $(x_n)_{n\geq 1}$ is a sequence in $X$ converging weakly to $x\in X$, then $T\in \mathcal{K}(X,Y)$.
Then, the hint says,
Suppose that $T$ is not compact. Show that there exists $\epsilon>0$ and a sequence $(x_n)_{n\geq 1}$ in the unit ball of $X$ such that $\| Tx_n - Tx_m\|\geq \epsilon$ for all $n\neq m$. Then show that $(x_n)_{n\geq 1}$ has a weakly convergent sequence.
Edit2: This is similar to this post, where one may put $X=E$ in the answer there. On the other hand, I'd very much like to know how to show that there exists $\epsilon>0$ and a sequence $(x_n)_{n\geq 1}$ in the unit ball of $X$ such that $\| Tx_n - Tx_m\|\geq \epsilon$ for all $n\neq m$. I have explained what I thought in the comment below.
You should proof it by contradiction as the hint suggests it. If the hint is satisfied then you have found a weakly convergent (sub-)sequent $(x_{n_k})_{k \in \mathbb{N}}$ for which $(T(x_{n_k}))_{k \in \mathbb{N}}$ isn't Cauchy, thus not convergent which contradicts your conditions.