Is such an infinite dimensional metric space, weakly contractible?

We counteract this answer by adding the rigidity assumption: Is there still a counterexample?

Let $(X_{n},d_{n})_{n \in \mathbb{N}}$ be a sequence of complete geodesic metric spaces such that:
$X_{n}$ is a regular$^1$ CW-complex of constant local dimension$^3$ $n$, it is of finite type$^4$, boundaryless$^2$, unbounded, uniform$^5$, and it is the $n$-skeleton of $X_{n+1}$, which is n-connected. Moreover, the distances $d_{n}$ , $d_{n+1}$ generate the same topology on $X_{n}$ and $\forall x,y \in X_{n} \ d_{n+1}(x,y) \le d_{n}(x,y)$.
Finally $(X_{n},d_{n})$ is quasi-isometric to $(X_{n+1},d_{n+1})$, through the inclusion map $X_{n} \subset X_{n+1}$, and a distance $d$ on $ \bigcup{X_{n}}$ is defined (for $x, y \in X_{n_0}$) by $d(x,y) := lim_{n (\ge n_0) \to \infty} d_{n}(x,y)$.

Rigidity assumption: if $S$ is a connected subspace of $X_{n}$ such that $S$ contains the geodesic paths between all its points, for $d_{n}$ and $d_{n+1}$, then $d_{n} = d_{n+1}$ on $S$.

Definition : Let $X:=\overline{\bigcup{X_{n}}}$ be the completion of the metric space $\bigcup{X_{n}}$ with $d$.
Question : Is $X$ weakly contractible ?

Remark: Some of these conditions could be useless for a proof, and others, highly generalized.
Motivation: See here for applications to geometric group theory and noncommutative geometry.

$^1$Regular (for a CW complex) : the attaching maps are homeomorphism (see this post).
$^2$Boundaryless (for a regular CW complex) : the boundary of each closed cell is contained is the union of the boundaries of other closed cells.
$^3$Constant local dimension : the topological dimension of all neighborhood of all point, is constant.
$^4$Finite type : finitely many $r$-cells ending in a fixed $(r-1)$-cell.
$^5$Uniform : For all $r$-cell $c_{1}$ and $c_{2}$, there is a neighborhood $n_{1}$ of $c_{1}$ and $n_{2}$ of $c_{1}$, such that $n_{1}$ is homeomorphic to $n_{2}$.

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