Homotopy problem for infinite dimensional topological space II 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.
 A: The following does not settle the original question.
The answer is `no' if we ignore the intrinsic metric requirement and merely demand metric.
For a counterexample first observe that the inverse limit $X$ of nested closed intervals $X_{n}$ (with retract bonding maps) might not be path connected. 
(For example if $X$ is the closure of the graph of y= sin(1/x) with x in [0,1], and for $X_{n}$ take the subspace with x in [1/n,1]).
Now to arrange the correct dimensions multiply each $X_{n}$ by an n-1 cube (canonically embedded in Hilbert space) and adjust the bonding maps to collapse the terminal coordinates of the thickened $X_{n}$ factors.

The category of compact metric spaces X, each of which is the inverse limit of nested finite dimensional contractible retracts $X_{n}$, provides a more general source of examples.
Most of the conditions in the questions are satisfied automatically for such X, since any two compact metric spaces are quasi isometric, and the inverse limit topology is compatible with the metric as described in the question.
Multiplying factors by n-cubes arranges for dimension going to infinity.
As sketched in the comments, we can arrange a counterexample X with nontrivial fundamental group, by compactifying the open unit disk with an annulus, while remaining in the category at hand.
A: 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).
