In the paper, "Banach Spaces which can be given an equivalent uniformly convex norm" by Per Enflo, there is one lemma whose proof I failed to understand:
Lemma. If a Banach Space $B$ does not have the finite tree property, then for every $\varepsilon>0$ there is an $n$ and a $\delta>0$, such that if $z \in B$ and $(x_1,x_2,\dots,x_{2^n})$ is an $(n,\varepsilon)$-partition of $z$ then $$\sum_{ j=1}^{2^n} \|x_j\| \ge (1+\delta)\|z\|.$$
Definitions:
An ordered pair $(x_1,x_2)$ in a Banach space is a $(1,\varepsilon)$-part of a tree if $\|x_1 - x_2\| > \varepsilon$. Now if the $(n,\varepsilon)$-part of a tree is defined, we say that a $(2^{n+1})$-tuple $(x_1, x_2,\dots,x_{2^{n+1}})$ is a $(n + 1, \varepsilon)$-part of a tree if $\|x_{2j-1} -x_{2j} \| > \varepsilon$, for $j=1,\dots,2^n$ and the $2^n$-tuple $((x_1 + x_2)/2, (x_3 + x_4)/2, \dots)$ is an $(n,\varepsilon)$-part of a tree.
The Banach space $B$ has the finite tree property (FTP), if there is an $\varepsilon > 0$ such that for every n, there is an $(n,\varepsilon)$-part of a tree where all elements have norm at most $1$.
The ordered pair $(x_1,x_2)$ is a $(1,\varepsilon)$-partition of $z$ if $$x_1 + x_2 = z, \|x_1\|=\|x_2\|, \ and \ \|x_1/\|x_1\| - x_2/\|x_2\| \| \geq \varepsilon.$$ Having defined an $(n, \varepsilon)$-partition of $z$, we say that the $2^{n+1}$ tuple $(y_1, y_2,..., y_{2^{n+1}})$ is an $(n+1,\varepsilon)$-partition of $z$ if $\|y_{2j-1}\|=\|y_{2j}\|$ , $\|y_{2j-1}/\|y_{2j-1}\| - y_{2j}/\|y_{2j}\| \| \geq \varepsilon$ for $1\leq j \leq2^n$ and the $2^n$-tuple $(y_1+y_2,y_3+ y_4,\dots)$ is an $(n,\varepsilon)$- partition of $z$.
The proof is rather short, it says assume $z$ has norm equal to 1.Taking a $(n,\varepsilon)$-partition of $z$ and multiplying by $2^n$ every element, we get a $(n,\varepsilon)$-part of a tree whose vectors have all norm $\geq 1$.(It is easy to prove by induction that all elements in a $(n, \varepsilon)$-partition of $z$ have norm at least $ 1/2^{n}$). Now since $B$ does not have the FTP there is an $n$ and a $\delta>0$ such that for all $(n,\varepsilon)$-parts of trees formed by multiplication by $2^n$ of the vectors in a $(n,\varepsilon)$-partition, there is one vector of length $\geq 1+2^n \cdot \delta$ and then the result follows.
My question is, that $\delta$ depends on the chosen partition and so that delta doesn't necesarily have to work for all partitions. I can't see if there is something I am missing.