Weakly quasisymmetric maps of a connected doubling space are quasisymmetric I'm currently reading through a few chapters of Juha Heinonen's Lectures on Analysis on Metric Spaces, and I'm having some trouble understanding the finer points of a particular proof. The result is located in Chapter 10 on quasi-symmetric maps:

Theorem 10.19. A weakly quasisymmetric embedding of a connected doubling space is quasisymmetric, quantitatively.

Here is the first bit of the proof, which is what I'm having trouble with. 
Proof. Let $X,Y$ be doubling metric spaces and, in addition, $X$ is connected. Let $$t = \frac{|x-a|}{|x-b|}, \hspace{0.5in} t' = \frac{|f(x)-f(a)|}{|f(x)-f(b)|}, $$
where $|x-a|$ denotes the distance from $x$ to $a$ in $X$. We show $t' \leq \eta(t)$, where $\eta(t) \to 0$ as $t \to 0$.
The first case is when $t > 1$ (this is the case I am having difficulty following). Denote by $C_{1}$ the covering function of $X$. Set $r = |b-x|$ and $$\varepsilon = \min \left\{ \frac{r}{2 C_{1}\left(\frac{1}{6t} \right)}, \frac{r}{2} \right\}.$$ Join $x$ to $a$ by an $\varepsilon$-chain $x = x_{0}, \ldots , x_{N} = a$. Choose a subsequence $a_{0}, a_{1}, \ldots$ of $(x_{i})$ so that $a_{0}=x_{0}=x$ and that $a_{i+1}$ is the last point $x_{j}$ with $|x_{j} - a| \leq r - i \varepsilon$. Because $\varepsilon \leq r/2$, the point $a_{i+1}$ is well-defined with $a_{i+1} \neq a_{i}$ whenever $\{a_{0}, \ldots, a_{i} \} \subset B(x,tr)$ and $i \varepsilon \leq r/2$.We show that under these conditions, we have that $(i+1)\varepsilon \leq r/2$.
After this is shown, he goes on to say:
It follows that there is an integer $s$ so that the points $a_{0}, \ldots, a_{s+1}$ are well defined, $|a_{j} - x|<tr$ for $0 \leq j \leq s$, and $|a_{s+1} - x| \geq tr = |a-x|$.
I am confused as to what the subsequence $(a_{i})$ is supposed to be. My thinking is that the $\varepsilon$-chain could, a priori, wander outside $B(x,tr)$, and this refinement is a way of controlling how far away the points in the sequence wander from $x$. I am also somewhat confused as to why the proof that $(i+1)\varepsilon \leq r/2$ is necessary, and how this fact gives that $a_{0}, \ldots, a_{s+1}$ are well-defined. If anyone familiar with the proof or the argument could help, I would appreciate it!
 A: The goal is to construct a sequence $a_0=x, a_1,\dots,a_{s+1}$ so that 


*

*$|a_{i+1}-a_i|\le |a_i-a_{i-1}|$ for all $i$

*$|a_{s+1}-x|\ge tr$

*The length of sequence is bounded by a number independent of $a,b,x$ (it may depend on $t$). 


Property 1 yields $|f(a_{i+1})-f(a_i)|\le H|f(a_i)-f(a_{i-1})|$ for every $i$. Applying this repeatedly and using #3, we get a bound on $|f(a_{s+1})-f(x)|$. Then the property 2 gives a bound  on $|f(a)-f(x)|$, since $|a-x|=tr$. So, the rest of the proof is the construction of a sequence with properties 1,2,3.
A way to achieve #1 is to somehow choose the points  so that 
$$
r-(i+1)\varepsilon \le |a_{i+1}-a_i| \le r-i\varepsilon \tag{4}
$$
for every $i$. To this end the author first takes a sequence $x_j$ with $|x_{j+1}-x_j|\le \varepsilon$ and then refines it by picking $a_{i+1}$ as 


*

*the last $x_j$ element with $|x_j-a_i|\le r-i\epsilon$. 


Then both inequalities in (4) hold: the upper bound is built-in, while the lower follows from the fact that if $|x_j-a_i|<r-(i+1)\varepsilon $, then 
$$|x_{j+1}-a_i|<\varepsilon+r-(i+1)\varepsilon = r-i\varepsilon$$
a contradiction the word last in the bullet item.
The  sequence will go on until it reaches $|a_{i}-x|\ge tr$; then it stops. This fulfills property 2.
There are still issues to resolve. For $a_{i+1}$ to be defined, we need some element $x_j$ with $|x_j-a_i|\le r-i\varepsilon$. To ensure this, it suffices to show that $r-i\varepsilon\ge \varepsilon$, because the points $x_j$ form an $\varepsilon$-chain. Since $\varepsilon\le r/2$ by choice, it's enough to have $i\varepsilon\le r/2$. 
The doubling condition is used to show that $i\varepsilon\le r/2$ by induction: if $i\varepsilon\le r/2$ holds, then the selected sequence $a_0,\dots,a_i$ is separated  at least by $r/2$, and since it's contained in the ball $B(x,tr)$, its cardinality is controlled by the doubling condition; from there one gets $(i+1)\varepsilon\le r/2$.
By the way, the last paragraph also ensures property 3: the length of the sequence is at most $r/(2\varepsilon)$, and since $\varepsilon$ was chosen  proportional to $r$, this ratio is bounded independently of $r$.
