Non strictly singular operators

Non strictly singular operators

Let $X$ be a separable Banach space and let $T:X\to X$ be a bounded
operator that is not strictly singular. Can we always find an infinite
dimensional, closed, and complemented subspace $Y$ of $X$ such that
$T:Y\to T(Y)$ is an isomorphism?
I cannot show this is true, and I cannot think of a counterexample. The
only idea of counterexample I had was looking at HI spaces, where every
operator is of the form $\lambda I+S$, where $S$ is strictly singular.
However, as far as I know, such operators (when $\lambda\neq 0$) are
isomorphisms on a closed, finite-codimensional subspace, thus
complemented.