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In paper "Pointwise products of uniformly continuous functions" by Sam B. Nadler, Jr., He defined the finitely chainable as followings :

Let $(X,d)$ be a metric space. An $\varepsilon$-chain in $X$ from $x$ to $y$ of length $m$ is a finite sequence $$ x_0 =x,x_1,x_2, \dots, x_m = y $$ such that $d(x_{i-1},x_i) < \varepsilon$ for all $i = 1,2, \dots, m $. We say that $(X,d)$ is finitely chainable provided that for each $\varepsilon > 0$, there are finitely many points $p_1,p_2, \dots , p_n \in X$ and a positive integer $m$ such that there is an $\varepsilon$- chain in $X$ of length $m$ from any point $x$ of $X$ to one of the points $p_1,p_2 , \dots , p_n$.

Why open unit ball in any infinite dimensional Banach space is finitely chainable? Note that he hinted by using points on radial lines from the center of open unit ball to form the required chains.

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It is enough to show that $B(0,1)=\{x\in X : ||x||<1\}$ is finitely chainable. Take any $\varepsilon >0$ and let $m=\left[\frac{1}{\varepsilon}\right] +1$ and let $x\in B(0,1).$ Then there exists a chain from $x$ to $0$ of lenght $m .$ To see this it is enough to consider points $x_1 =0, x_2 =\frac{1}{m+1} x ,..., x_{m-1} =\frac{m-1}{m+1} x , x_m =\frac{m}{m+1} x , x_{m+1} =x .$

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