Selecting points from an increasing sequence of $\varepsilon_n$-separated-covering, for any $x$, can we find an infinite chain converging to $x$? Let $(X,d)$ be a metric space and denote by $\mathbb{N}$ the set of non-negative integers $\{0,1,2,\dots\}$.
Suppose that $(\varepsilon_{n})_{n \in \mathbb{N}}$ is a strictly decreasing sequence of positive real numbers such that $\varepsilon _n \to 0, n\to \infty $.
Suppose also that $X_0\subset X_1 \subset X_2 \subset \dots\subset X$ are such that

*

*$\forall n \in \mathbb{N}, \forall x,x' \in X_n, \qquad (x\neq x')\implies d(x,x')> \varepsilon_n \qquad$ (i.e., $X_n$ is $\varepsilon_n$-separated with respect to the metric $d$)

*$\forall n \in \mathbb{N}, \forall x\in X, \exists x_n \in X_n, \qquad d(x,x_n)\le \varepsilon_n \qquad$ (i.e., $X_n$ is a $\varepsilon_n$-cover of $X$ with respect to the metric $d$)

Fix any $x \in X$.
If $n \in \mathbb{N}$, pick $x_n \in X_n$ such that $d(x,x_n) \le \varepsilon_{n}$, then pick $x_{n-1} \in X_{n-1}$ such that $d(x_n,x_{n-1})\le \varepsilon_{n-1}$, and continue until reaching the index $0$. This way, we have obtained a string $x_0 \in X_0,x_1 \in X_1,\dots,x_n \in X_n$ such that
$$d(x_1,x_0)\le \varepsilon_0, \; d(x_2,x_1) \le  \varepsilon_1,\; \dots,\; d(x_n,x_{n-1}) \le \varepsilon_{n-1},\; d(x,x_n) \le \varepsilon_{n}$$
Notice that this construction relies on the initial knowledge of $n$ and it is built working backward, starting from $n$ and going back to $0$.
Now, I'm not quite satisfied with this construction, and I'm wondering if we can do "nearly the same" in general without this prior knowledge of $n$, i.e.,

Can we find a sequence $(x_n)_{n \in \mathbb{N}}$ such that

*

*$\forall n \in \mathbb{N}, x_n \in X_n$

*$\forall n \in \mathbb{N}, d(x_{n+1},x_{n}) \le \varepsilon_{n}$

*$d(x_n,x) \to 0, n \to \infty\quad$?


If this is possible, can we strengthen somehow condition $3$ obtaining some convergence rate depending on the sequence $(\varepsilon_n)_{n \in \mathbb{N}}$ (i.e., replace condition 3 with something stronger like "there exists a universal constant $c$ such that $\forall n \in \mathbb{N}, d(x_n,x) \le c \varepsilon_n$")?
I've tried to work forward instead of backward, but this doesn't seem to work, at least if we do it in a naive way: for example, if we proceed selecting points in a greedy way (i.e., assuming that each $X_n$ is finite, if we inductively pick the closest point to $x$ from $X_n$ that satisfies the constraint 2) we can end up to a dead end where the construction can't go further).
 A: It's false in general. Here a counterexample.
Let $\varepsilon_n:=3^{-{n}}$.
Let $H$ be an infinite dimensional separable Hilbert space with $(e_n)_{n \in \mathbb{N}}$ as a Hilbert base.
Consider $Y_0 := \{\varepsilon_0 e_n\}_{n \ge 0}, \;Y_1 := \{\varepsilon_1 e_n\}_{n \ge 1},\; \dots,\; Y_k := \{\varepsilon_k e_n\}_{n \ge k},\; ...$.
Define $X:= \{0\} \cup \bigcup_{k \in \mathbb{N}} Y_k$ and let $d$ the metric on $X$ induced by the Hilbert norm.
Let $X_0 := Y_0,\; X_1:=Y_0\cup Y_1,\; \dots,\; X_k = Y_0\cup Y_1\cup \dots\cup Y_k,\;\dots$.
It's clear that $X_n$ is $\varepsilon_n$-separated (using Pythagoras theorem for the points along two different coordinates and by inspection along points on the same coordinate) and that it is a $\varepsilon_n$-cover.
However, if $(x_n)_{n \in \mathbb{N}}$ is a sequence such that

*

*$\forall n \in \mathbb{N}, x_n \in X_n$

*$\forall n \in \mathbb{N}, d(x_{n+1},x_{n}) \le \varepsilon_{n}$
then, using the fact that if $i,j  \in \mathbb{N}$ are such that $i \neq j$ then for each $m,n \in \mathbb{N}$ we have that $d(\varepsilon_me_i,\varepsilon_ne_j) =\sqrt{\varepsilon_m^2 + \varepsilon_n^2} > \min(\varepsilon_m,\varepsilon_n)$, we can deduce that if $i \in \mathbb{N}$ is such that $x_0 = \varepsilon_0 e_i$, then $x_1 \in \{\varepsilon_0e_i,\varepsilon_0e_i\}$ (for any element $x'\in X_1$ different from $x_0$ and $\varepsilon_1 e_i$ we have that $d(x_0,x')>\varepsilon_1$), i.e., $x_0$ and $x_1$ shares the same direction $e_i$. In the same manner, $x_2$ has to share the same direction $e_i$ of $x_0,x_1$ and so forth. Overall, we obtain that there exists a direction $e_i$ such that $\forall n \in\mathbb{N}, x_n$ belongs to the set $\{ \lambda e_i \mid \lambda >0\}$. But for each coordinate $j$, there exists only $j+1$ different element in $X$ which are in the direction of $e_j$, i.e. $\varepsilon_0 e_j, \dots, \varepsilon_j e_j$. It follows that there exists a coordinate $i$ such that $\forall n \in \mathbb{N}, x_n \in \{\varepsilon_0 e_i, \dots, \varepsilon_i e_i\}$, which implies that $d(x_n,x)$ can't converge to $0$ as $n \to \infty$.
