TITLE: Homotopy problem for infinite dimensional topological space II
QUESTION [5 upvotes]: This post here is a specification of this post.
Let $(X_{n},d_{n})_{n \in \mathbb{N}}$ be a sequence of intrinsic metric spaces verifying :

$X_{n}$ have topological dimension $n$. 
$X_{n+1}$ is n-connected.
$X_{n} \subset X_{n+1}$
The distance $d_{n}$ and $d_{n+1}$ generate the same topology on $X_{n}$. 
$\forall x,y \in X_{n}$ : $d_{n+1}(x,y) \le d_{n}(x,y)$.
$(X_{n},d_{n})$ is quasi-isometric to $(X_{n+1},d_{n+1})$

Definition : Let $d$ be a distance on $ \bigcup{X_{n}}$, defined as follows :   $d(x,y) = lim_{n \to \infty} d_{n}(x,y)$.   
Remark : There is a small abuse in the previous definition because $d_{n}(x,y)$ is defined only for $x, y \in X_{n}$. But because we take $n \to \infty$, there is no problem.
Definition : Let $X:=\overline{\bigcup{X_{n}}}$, the complete metric space obtained as a completion of $\bigcup{X_{n}}$ for $d$.

Question : Is $X$ weakly contractible ?

Remark : If yes, perhaps some of these conditions are useless in the proof, and perhaps the useful conditions can be highly generalized.

REPLY [2 votes]: The intristic metric assumption seems to be added to exclude $S^1$ from the previous post. But what about $S^d$? It can be approximated with bigger and bigger disks
$\{x\in S^d\ |\ x_1\leq 1-\frac{1}{n} \}$ (with the natural intristic metric) and in the limit you get the sphere. (Of course dimension of such a disk is $d$ so one needs some obvious bypasses for $n<d$.)
As it was already noticed quasi-isometry requirement is trivial for bounded spaces.
Similarly with the dimension assumption: we can always take cartesian product with some nice space like $\prod_{i=1}^n [0,\frac{1}{2^i}]$ with the $\ell_2$-metric ($\frac{1}{2^i}$ is unnecessary but this way all the bounds are uniform).