Suppose ($x_n$) is a normalized, linearly independent, sequence in a reflexive Banach space $X$, and $T$ is an injective, strictly singular, bounded operator on $X$ such that $Tx_n\longrightarrow 0$. Does there exist a subsequence $(y_n)$ of $(x_n)$ such that $T$ restricted to the closed span of $(y_n)$ is compact? When $(x_n)$ is a basic sequence and $\sum||Tx_n||$ converges (so $Tx_n\longrightarrow 0$ fast enough), it is not hard to show that $T$ restricted to closed span of $(x_n)$ is indeed compact (without any need to assume that $T$ is strictly singular or $X$ is reflexive).
An operator is called strictly singular if it is not an isomorphism when restricted to any infinite dimensional subspace. Compact operators are always strictly singular but not the other way around. However, strictly singular are compact in $l_p$ and $c_0$.