A possible characterization of WSC spaces

A Banach space $$X$$ is weakly sequentially complete (WSC) if every weakly Cauchy sequence in $$X$$ is weakly convergent.

I will use the following classical result:

Rosenthal's $$\ell_1$$ theorem: Every bounded sequence in a Banach space $$X$$ either contains a weakly Cauchy subsequence, or a subsequence equivalent to the $$\ell_1$$ basis.

It follows from the fact that a closed subspace of a WSC space is WSC, that every subpace of a WSC space is either reflexive or contains $$\ell_1$$. This observation, if localized, actually gives us a characterization of WSC spaces, namely:

Proposition. For a Banach space $$X$$ TFAE:

1. $$X$$ is WSC;
2. Every bounded sequence in $$X$$ has either a weakly convergent subsequence, or a subsequence equivalent to the $$\ell_1$$ basis;
3. Every weakly precompact (i.e. not containing a sequence equivalent to the $$\ell_1$$ basis) bounded set in $$X$$ is relatively weakly compact.

It would thus feel plausible that the "nonlocalized" version of this Proposition could be true as well. Namely, I would like to know if:

Question: Let $$X$$ be a Banach space such that every closed subspace $$Y$$ of $$X$$ is either reflexive or contains $$\ell_1$$, is $$X$$ then weakly sequentially complete?

As mentioned above, the converse implication is true. I have, however, not managet to prove/disprove the Question above. Any help will be appreciated.

The article by Azimi & Hagler provides a Banach space $$X$$ with a basis $$(e_n)$$ such that
• $$X$$ is hereditarily $$\ell^1$$ : every infinite-dimensional closed subspace of $$X$$ contains a copy of $$\ell^1$$.
• $$(e_n)$$ is weakly Cauchy but not weakly convergent.
• Thank you for the answer! I have some questions: 1. How do we know that $T$ is weakly precompact? 2. How do we know that $T$ is onto $Y$ (or at least that its range contains $Z$)? 3. What is the need to define $V$, cannot we just take $V=\ell_1$ and proceed in the smae way? Thanks again! Sep 20 at 12:37
• @KeeperOfSecrets You spotted a mistake in your (2.) Please see the edited answer. It turns out that there exist non-wsc hereditarily $\ell^1$ Banach spaces. Sep 20 at 15:17