Let me first fix some notations.

If $A$ and $B$ are nonempty subsets of a Banach space $X$, we set $$d(A,B)=\inf\{\|a-b\|:a\in A,b\in B\},$$$$\widehat{d}(A,B)=\sup\{d(a,B):a\in A\}.$$

Let $A$ be a bounded subset of a Banach space $X$. The Hausdorff measure of non-compactness of $A$ is defined by $$\chi(A)=\inf\{\widehat{d}(A,F):F \subset X\},$$ where the infimum is taken over all finite subsets $F$ of $X$. Clearly, $A$ is relatively compact if and only if $\chi(A)=0$.

The following result is well-known (see Proposition 2.c.4 in the book of J. Lindenstrauss and L. Tzafriri):

Let $X,Y$ be infinite-dimensional Banach spaces. Assume that an operator $T: X\rightarrow Y$ is such that the restriction of $T$ to any subspace of finite co-dimension in $X$ is not an isomorphic embedding. Then, $\forall \epsilon>0$, there exists an infinite-dimensional closed subspace $M$ such that $T|_{M}$ is compact and $\|T|_{M}\|<\epsilon$.

I am thinking about the quantitative version of this result.

Question: Let $X,Y$ be infinite-dimensional Banach spaces and let $c>0$. Assume that $T: X\rightarrow Y$ is an operator such that $\|T\|=1$ and $\chi(TB_{M})>c$ for every infinite-dimensional closed subspace $M$ of $X$. Is there an infinite-dimensional closed subspace $X_{0}$ of $X$ such that $\|Tx\|\geq c\|x\|$ for all $x\in X_{0}$?

Thank you!