Are small balls in a metric space quasi-symmetric to a compact ball connected? I'm trying to solve a problem but I'm stuck. Maybe someone can help me.
I will denote every metric on every metric space by $d$ and I will use closed balls, denoting them with the letter $B$, so for $x\in X$ and $r>0$
$$B(x,r)\colon=\lbrace y\in X\:\colon \: d(x,y)\leq r\rbrace.$$ To the problem:
let $X$ be a metric space and let $V$ be a compact ball inside a Carnot group $G$, endowed with the usual CC metric. Assume that there exists a quasi-symmetric homeomorphism $f\colon V\longrightarrow X$.
Question: Are small enough balls in $X$ connected? That is, does there exists a constant $K>0$ such that all balls of radius $\leq K$ in $X$ are connected?
Intuitively I think that the answer should be yes: $V$ is a closed ball inside a Carnot group and therefore is compact, connected, simply connected i.e. very nice. $X$ not only is homeomorphic to $V$ but it's also quasi-symmetrically equivalent to it. It cannot be that ugly, right? All the "bad cases" I can think of do not have compact boundary.
I tried using that quasi-symmetric maps on uniformly perfect spaces are $\alpha$-Hölder continuous for some $\alpha\in (0,1]$:
$\textbf{Lemma:}$ Suppose that $X$ is a bounded and uniformly perfect metric space. Let $f\colon X\longrightarrow Y$ be a quasisymmetric homeomorphism. Then there are constants $A,B\geq 1$ and $\alpha\in (0,1]$ such that 
$$ \frac{1}{A}d(x,y)^{1/\alpha}\leq d(f(x), f(y))\leq Bd(x,y)^\alpha.$$
Since connected spaces are uniformly perfect we can apply this to our situation to obtain that for every closed ball $B(x,r)$ there is a connected set $U$ and a constant $D$ such that
$$ B(x,\frac{1}{D}r^{1/\alpha^2}) \subset U\subset B(x,r).$$ 
I'm not sure this is the right approach as I can't make any progress from here. I'm aware that since the metric space is compact, it would be enough to show that the closure of each open ball is the closed ball with the same radius, in order to say that balls are connected. But I couldn't show this and I don't even expect every ball to be connected..
Does anyone have other ideas that I can try? Thank you very much!
 A: This is not true even in the simplest case when $G=\mathbb{R}$ and $f$ is not just quasisymmetric but bi-Lipschitz.
I will take my ball to be the interval $[0,1]$ in $\mathbb{R}$ and take $X$ to be the graph of a Lipschitz function $g:[0,1]\rightarrow\mathbb{R}$. Then then the downward projection from $X$ (the graph of $g$) to $[0,1]$ is the bi-Lipschitz map $f$.
Now I define my Lipschitz function $g$ as follows:


*

*$g(2^{-n}) = 0$ for $n=0,1,2,\dots$ and $g(0)=0$.

*On each interval $[2^{-(n+1)}, 2^{-n}]$, $g$ makes a ``triangular tent'' of slope $4$.


So, e.g., $g(1/2)=0$, on $[1/2, 3/4]$ $g$ rises with slope 4 so that $g(3/4)=1$, then on $[3/4,1]$ $g$ descends with slope $4$ back to zero. It does the same thing (scaled down) on each dyadic interval.
It's clear that $g$ is lipschitz with lipschitz constant $4$.
Let $X$ be the graph of $g$ (with the ambient metric from the plane).
For each $n$, consider the closed ball $\overline{B}((2^{-n},0),2^{-(n+1)})$. None of these are connected, since they each contain the two "base points" of the tent to their left, but not the "peak" of the tent.
For example, $\overline{B}((1,0), 1/2))$ contains $(1/2,0)$ and $(1,0)$, but these are in different connected components of the ball, since the ball does not contain the peak $(3/4,1)$.
(I think pictures would make this clearer but I'm not good at drawing pictures without a lot of effort.)
On the positive end, I think for your original question one could say something like "$X$ is bi-Lipschitz equivalent to a space in which all sufficiently small closed balls are connected".
