$\newcommand{\span}{\operatorname{span}}$EDIT: By some further reading, I note that the very fact that $\{x_n\}$ does not contain a weak convergent subsequence implies by Rosenthal's $\ell_1$ theorem (which is precisely reference (B)!) that $\{x_n\}$ will contain a subsequence $\{x_n'\}$ identifiable with the Schauder basis of $\ell_1$. Assuming I take Rosenthal's theorem for granted, I still am confused as to how this implies $\{x_n'\}$ is a basis of its own closed linear span. That it is a basis of its own span is obvious, but as discussed below I don't know how that relates to the closure of this span in $E$.
OP:
The following lemma is very important to a proof I am following:
Let $E$ be a nonempty irreflexive Banach space. $E$ contains a nonempty closed subspace $X\subset E$ of infinite codimension, with $X$ also irreflexive.
The given proof is very brief, relying on two references:
By the Eberlein-Smulian theorem, there is a bounded sequence $\{x_n\}$ in $E$ with no weakly convergent subsequence, in particular [see (A) and (B)] it contains a basic subsequence $\{x_n'\}$; let $X$ be the closed linear span of $\{x_{2n}'\}\,\blacksquare$
Reference (B) is a bit hard for me to read; the pertinent bit seems to be the very beginning discussion, where it is stated:
A bounded sequence of elements $(f_n)$ of a Banach space is said to be equivalent to the usual $\ell_1$ basis provided there is a $\delta\gt0$ so that for all $n$ and all scalars $c_1,\cdots,c_n$: $$\delta\sum_{i=1}^n|c_i|\le\left\|\sum c_if_i\right\|$$"Of course", if $(f_n)$ has this property, then the closed linear span of the $f_n$ is isomorphic to $\ell_1$.
I believe from this I am supposed to infer that $(f_n)$ is a basic sequence, i.e. every element of $\overline{\span\{f_n:n\in\Bbb N\}}$ is expressible as a finite linear combination of the $f_n$, so we can show that there exists a $\{x_n'\}$ satisfying this, then because the closed linear span is isomorphic to $\ell_1$ I can conclude that $\{x_n'\}$ is a basic sequence. However, although the inequality shows that the linear map $J:\span\{f_n:n\in\Bbb N\}\to\ell_1$ characterised by $J:f_n\mapsto e_n$, where $e_n$ is the $n$th unit vector of $\ell_1$,is a bicontinuous map by norm equivalence (since $\|\sum c_if_i\|\le\sup\|f_i\|\sum |c_i|\lt\infty$ as $f_n$ is bounded) and as such an isomorphism: however, I have qualms with the notion that this isomorphism remains an isomorphism, or even that it remains well-defined, when the domain is adjusted to the closed linear span instead. The article treats this as obvious but I do not understand closed linear span very well.
My thoughts:
The closure of the linear span, $X$, of some $\{f_n\}$ is a linear subspace, since if $x\in X$, given an $\epsilon\gt0$, there is an $n\in\Bbb N$, scalars $c_1,\cdots,c_n$ and vectors $f_{k_1},\cdots,f_{k_n}$ such that $\|x-\sum c_if_{k_i}\|\lt\epsilon$; hence $\lambda x$ for any scalar $\lambda$ will satisfy by norm homogeneity $\|\lambda x-\sum (\lambda c_i)f_{k_i}\|\lt\lambda\epsilon$ and so $\lambda x$ is again in the closure of the span. A similar argument shows that if $u,v\in X$ so too is $u+v$.
However, there may exist $x\in X$ but $x$ is not in $\span\{f_n:n\in\Bbb N\}$. Thus, one cannot argue that $f_n$ remains a basis of $X$ unless $\span\{f_n:n\in\Bbb N\}$ is itself closed. Then the discussion of reference (B) confuses me, as I do not see how to extend the definition of $J$ so that the isomorphism holds; indeed, even if one could extend $J$, we surely still could not argue that $f_n$ is a basis of the closed span by my thoughts above. I am missing the point - I do not even understand the fascination with being isomorphic to $\ell_1$, and why it is relevant.
Secondly, reference (A) gives a nice proof that for any bounded sequence $\{y_n\}$ with no subsequence converging to $0$ (it actually states $0\lt\inf_n\|y_n\|$, but I believe the two are equivalent), and any "norming set $G$ such that $g(y_n)\to 0,\,\forall g\in G$", where $G$ is a norming set if $G\subset Y^\ast$, and, where $S_G=\{g\in G:\|g\|=1\}$, for any $y\in Y$ one has $\sup_{g\in S_G}|g(y)|=\|y\|$, then there exists a subsequence $\{y_{n_k}\}$ such that for all scalars $t_i$ and any $q\lt p\in\Bbb N$, one has:
$$\delta\left\|\sum_{i=1}^qt_iy_{n_i}\right\|\le\left\|\sum_{i=1}^pt_iy_{n_i}\right\|$$
I note that this is similar but not quite the same as the condition assumed in the beginning discussion of reference (B).
Returning to the actual conditions in which I came across all this, the bounded sequence $\{x_n\}$ in question has no weakly convergent subsequence by definition, so I do not think one can choose a norming set $G\subset E^\ast$ such that $g(x_n)\to 0$ for all $g\in G$. As mentioned, even if one could, and one could use the result of reference (A), I would still be at a loss as to how this implies the existence of a basic subsequence.
As the reader can probably tell, I do not know much of this infinite-dimensional linear space theory; I am predominantly learning out of Royden's real analysis, but the proof I am following is a bit more advanced than this book (I pursue it because the result is very interesting to me even if the proof is hard) and I am lost in the weeds of "closed linear span"...
Many thanks to anyone who can clarify any of this.