 Solution

Assume weakly null sequences have norm-null images. Let $(y_n)$ be any sequence in $T(K)$, where $K$ is weakly compact, and choose $x_n\in K$ with $y_n=T{x}_n$. By the weak subsequence result, some $x_{n_j}\rightharpoonup x\in K$. Then $x_{n_j}-x\rightharpoonup 0$, and the assumed property givesThus every sequence in $T(K)$ has a norm-convergent subsequence with limit in $T(K)$. In a metric space, sequential compactness is equivalent to compactness. Therefore $\boxed{T(K)\text{ is norm compact}}$. This is one characterization of a completely continuous operator.