I will attempt to explain why a closed subspace generated by uncountable many branches intersects the subspace $X_*$.

First let's observe that for every $x= \sum_{i=1}^n \alpha_i\sigma_i $, with $ \{\sigma_i \}$ distinct branches, we have that $dist(x,X_*) \leq 2 max\{ |\alpha_i| : i=1...n\} $.

Let $A$ be an uncountable set of branches and assume that no element of $A$ is an isolated point.

Lemma: Let $ x = 1/n\sum_{i = 1}^{n} \sigma_i $ be an average of distinct branches in $A$.Then for every $\varepsilon > 0$ there exists $y$ in the linear span of $A$ such that: (i) $\|y\| \leq 4/n $ , (ii) $x+y$ is an average of branches and (iii) $dist ( x+y, X_* ) < \varepsilon$.

It is clear that if we have the above Lemma then by induction we could produce a sequence $(x_n)$ such that their norms are summable and $dist ( \sum_{j=1}^{n} x_j , X_* )\rightarrow 0$. Hence $\sum_{n=1}^ {\infty} x_n $ belongs to $X_*$.

The proof of the Lemma is a multiple application of the following observation.

Given $\sigma \in A$ and $k$ natural number. then for every $n$ there are $\{ \sigma_i \}_{i=1} ^ {n} $ distinct branches each one coinciding to $\sigma$ in the first $k$ of its elements. Set $ y = 1/n \sum _{i = 1} ^ {n} \sigma_i - \sigma $. Then $ \|y \| \leq 2$ , $x+y $ is an average and $dist ( y , X_* ) \leq 2/n$.

I understand that my explanation is rather incomplete but I hope that gives an idea of the approach.

I will attempt to explain why a closed subspace generated by uncountable many branches intersects the subspace $X_*$.

First let's observe that for every $x= \sum_{i=1}^n \alpha_i\sigma_i $, with $ \{\sigma_i \}$ distinct branches, we have that $dist(x,X_*) \leq 2 max\{ |\alpha_i| : i=1...n\} $.

Let $A$ be an uncountable set of branches and assume that no element of $A$ is an isolated point.

Lemma: Let $ x = 1/n\sum_{i = 1}^{n} \sigma_i $ be an average of distinct branches in $A$.Then for every $\varepsilon > 0$ there exists $y$ in the linear span of $A$ such that: (i) $\|y\| \leq 4/n $ , (ii) $x+y$ is an average of branches and (iii) $dist ( x+y, X_* ) < \varepsilon$.

It is clear that if we have the above Lemma then by induction we could produce a sequence $(x_n)$ such that their norms are summable and $dist ( \sum_{j=1}^{n} x_j , X_* )\rightarrow 0$. Hence $\sum_{n=1}^ {\infty} x_n $ belongs to $X_*$.

The proof of the Lemma is a multiple application of the following observation.

Given $\sigma \in A$ and $k$ natural number. then for every $n$ there are $\{ \sigma_i \}_{i=1} ^ {n} $ distinct branches each one coinciding to $\sigma$ in the first $k$ of its elements. Set $ y = 1/n \sum _{i = 1} ^ {n} \sigma_i - \sigma $. Then $ \|y \| \leq 2$ , $\sigma + y $ is an average and $dist (\sigma + y , X_* ) \leq 2/n$.

I understand that my explanation is rather incomplete but I hope that gives an idea of the approach.