Let $X$ be a Banach space. An operator $T:X\to X$ is called strictly singular iff for any infinite dimensional subspace $Y\subseteq X,$ $T|_{Y}:Y\to T(Y)$ is not an isomorphism.

It is known that for $X=\ell_p,$ $1\leq p<\infty,$ an operator is strictly singular iff it is compact. Also $T:\ell_\infty\to\ell_\infty$ is strictly singular iff $T$ is weakly compact. Can someone provide me proofs for these facts? I could not really locate proofs of the above mentioned facts in literature.