Every infinite-dimensional Banach space not containing $l^1$ has a normalized, weakly null sequence. I'm trying to deduce the following:

Theorem: Let $X$ be an infinite dimensional Banach space with no subspace $Y\le X$ with $Y\cong l^1$. There exists $(x_n)\in X$ such that $\Vert x_n\Vert =1$ and $x_n\rightharpoonup 0$.

from Rosenthal's $l^1$ theorem:

Theorem: Let $X$ be a Banach space with no subspace $Y\le X$ with $Y\cong l^1$. For all $(x_n)\in X$ bounded, there exists a subsequence $(x_{n_k})$ weakly Cauchy i.e. for all $\xi\in X^*$, $(\xi(x_{n_k}))$ is convergent in $\mathbb C$.

My attempt: First, choose $(x_n)$ linearly independent, normalized with dual elements $(\xi_n)$ i.e. $\xi_i(x_j)=\delta_{i,j}$. We can do this inductively: given such $x_1,\cdots,x_n$ and $\xi_1,\cdots, \xi_n$, we can choose $x_{n+1}\in\cap_{i=1}^n\text{ker}(\xi_i)$ normalized since $\text{dim}(X)=\infty$. Let $\xi_{n+1}\in X^*$ be normalized, extending the map given by $x_i\mapsto \delta_{i,n+1}$ on $\text{span}(x_1,\cdots,x_{n+1})$ using Hahn-Banach. Extract a weakly Cauchy subsequence so without loss of generality, assume $(x_n)$ is weakly Cauchy. Let $Y:=\overline{\text{span}(x_n|n\in\mathbb N)}$ and $Y^*=\overline{\text{span}(\xi_n|_Y|n\in\mathbb N)}$.
Suppose for a contradiction that there exists $\xi\in X^*$ such that $\xi(x_n)\rightarrow\alpha\neq 0$. I claim that $\Phi:l^1\rightarrow Y$ defined by extending $e_n\mapsto x_n$ (which is clearly bounded and injective) is an into isomorphism, contradicting hypothesis that $X$ does not contain $l^1$. I've thought of three ways to prove this:

*

*Show that $\Phi$ is surjective and use open mapping theorem.


*Show that $\Phi$ is bounded below.


*Show that $\Phi^*$ is surjective.
and I wasn't able to prove any of these. Does anyone have an idea how to prove this?
Thanks in advance!

Since $X$ is infinite dimensional, there is a sequence $(x_n)\subset X$ with $\Vert x_n\Vert \le 1$ with $\Vert x_n-x_m\Vert >\frac{1}{2}$ by Riesz lemma. By Rosenthal's theorem, we have a weakly Cauchy subsequence $(x_{n_k})$. Set $y_n=\frac{x_{n+1}-x_n}{\Vert x_{n+1}-x_n\Vert }$. Then for all $f\in X^*$, $|f(y_{n_k})|\le 2|f(x_{n+1}-x_n)|\rightarrow 0$ since $ (x_{n_k})$ weak Cauchy.
