I'm trying to understand the proof that a subsymmetric basic sequence in a Banach space is either weakly null or equivalent to the unit vector basis of $l_1$. This question was already asked and answered in this post, and I think I understand the structure of the argument (though I wonder if it can be done without contradiction).
The part I'm stuck at is when we assume the basic sequence $(x_n)$ is not equivalent to the unit vector basis of $l_1$. Towards a contradiction, suppose that $(x_n)$ is not weakly null. Then no subsequence of $(x_n)$ is, so in particular, no subsequence of $(x_n)$ converges to $0$. So there is $m>0$ such that $m \le |\!|x_n|\!|$ for all $n$. The answerer to the other post then says here that we can assume $(x_n)$ to be seminormalised, i.e., there exists $C\ge 1$ such that $$1/C \le |\!|x_n|\!| \le C$$ for all $n$>. I believe the lower bound but not the upper bound. I think I can follow the rest of the proof, so it's really just this snag that I'm hitting.
Is this where we are supposed to use the hypothesis that $(x_n)$ is not equivalent to the unit vector basis of $l_1$? I don't think it's used anywhere else in the proof, and if it isn't, then I can't see why one can't just structure the argument as a ``Weakly null? Done. Not weakly null? Then equivalent to canonical basis of $l_1$.'' sort of thing.
I realise there are a lot of definitions here, so I'll list them below. Perhaps I have a definition wrong, and that is what is causing confusion.
Definitions. A (Schauder) basis is a sequence $(x_n)$ of vectors such that every $x$ in the Banach space $X$ has a unique expression as $$x = \sum_{n=1}^\infty a_n x_n.$$ A sequence that is a basis for its closed linear span is a basic sequence. Two bases $(x_n)$ of $X$ and $(y_n)$ of $Y$ are equivalent if the map taking $x_n\mapsto y_n$ for all $n$ extends to a linear homeomorphism from $X$ to $Y$. This is the same as saying that there exist constants $0<m\le M$ such that $$ m\Big|\!\Big| \sum_{n=1}^\infty a_n x_n\Big|\!\Big| \le \Big|\!\Big| \sum_{n=1}^\infty a_n y_n\Big|\!\Big| \le M\Big|\!\Big| \sum_{n=1}^\infty a_n x_n\Big|\!\Big|. $$ for all sequences $(a_n)$ of scalars.
A sequence $(x_n)$ is seminormalised if for some $C\ge 1$, we have $1/C\le |\!|x_n|\!| \le C$ for all $n$, and it is normalised if $C=1$. A sequence is weakly null if it converges to $0$ in the weak topology. And the unit vector basis of $l_1$ is given by $(e_n)$, where $e_n$ has a $1$ in the $n$th coordinate and zeroes elsewhere.