Can a metric subspace be completely covered by balls after a finite number of steps? Let $X$ me a metric space with distance $d$ and $A$ be a subspace of $X$.
Let $B_\varepsilon(x)$ be the open ball centered in $x$ with radius $\varepsilon$, i.e. $\{y\in X\mid d(x,y) < \varepsilon\}$.
Let $B_\varepsilon(S) = \displaystyle\bigcup_{x\in S} B_\varepsilon(x)$
Let $S_0:=\{x_i\}_{i\in I_0}$ be a set of points such that:

$x_i\in A\quad \forall i\in I_0$.
$B_\varepsilon(x_i)\cap B_\varepsilon(x_j) = \emptyset\quad \forall i\neq j$
If $x\in A\setminus S_0$ then $\exists i\in I_0$ such that $B_\varepsilon(x)\cap B_\varepsilon(x_i) \neq \emptyset$

(In other words, I'm taking a maximal set of points of $A$ so that any two are separated by $2\varepsilon$ or more)
At this point if $A\setminus B_\varepsilon(S_0) \neq \emptyset$ then we create:
$S_1:=\{x_i\}_{i\in I_1}$ such that:

$x_i\in A\quad \forall i\in I_1$.
$B_\varepsilon(x_i)\cap B_\varepsilon(x_j) = \emptyset\quad \forall i\neq j$
If $x\in A\setminus(S_0\cup S_1)$ then $\exists i\in I_1$ such that $B_\varepsilon(x)\cap B_\varepsilon(x_i) \neq \emptyset$

(In other words, I'm taking a maximal set of points of $A\setminus B_\varepsilon(S_0)$ so that any two are separated by $2\varepsilon$ or more)
At this point, if $A\setminus B_\varepsilon(S_0\cup S_1)\neq \emptyset$ then we create $S_2$ in a similar way. If $B_\varepsilon(S_0\cup S_1\cup S_2)$ does not cover $A$ we construct $S_3$ and so forth.
The questions are:
Given $\varepsilon > 0$ can we ensure that this process will finish after a finite amount of steps?
If so, is there an upper bound of said ampunt of steps?

A small example. Consider $X = A = \mathbb{R}$ with the euclidean distance.
Consider any $\varepsilon > 0$. Then let $k$ be the smallest integer such that $k\varepsilon \geq 1$. This means that $S_0$ has a countable amount of points.
We can order the points of $S_0$ acording to the usual total order $\leq$ of $\mathbb{R}$. In this case, $2\varepsilon < d(x_n,x_{n+1}) < 4\varepsilon$. In other words, the diameter of any conex subset of $\mathbb{R}\setminus B_\varepsilon (S_0)$ is less than $2\varepsilon$ (otherwise we could fit another ball in there)
In the next iteration we create $S_1 = \{y_i\}$ where $x_n < y_n < x_{n+1}$ and $\varepsilon < d(x_n,y_n) < 3\varepsilon$ and $\varepsilon < d(y_n,x_{n+1}) < 3\varepsilon$. In other words,  the diameter of any conex subset of $\mathbb{R}\setminus B_\varepsilon (S_0\cup S_1)$ is less than $\varepsilon$.
This means that when we put the points of $S_2$ in said connex subsets (one in each) then $\mathbb{R}\setminus B(S_0\cup S_1\cup S_2) = \emptyset$.
Thus three steps are sufficient to cover $\mathbb{R}$ with balls of any radius $\varepsilon$.
 A: For a metric space $(A,d)$ and $\varepsilon>0$ by $DG(A,d,\varepsilon)$ I denote the smallest number $N$ of steps needed to cover the $A$ for any realization of your (non-deterministic) algorithm and $\infty$, if there is no such $N$.
Remark. This definition can be naturally generalized to $DG(A,\mathcal U, U)$, where $(A,\mathcal U)$ is a uniform space and $U\in\mathcal U$ is an entourage.
Proposition 1. Let $(A,d)$ be the space $\Bbb R^n$  endowed with the standard metric and $\varepsilon>0$. Then $DG(A,d,\varepsilon)\le 5^n+1$.
Proof. Assume that after $N$ steps of the algorithm a point $x\in A$ left uncovered. Let $0\le i\le N-1$. By the maximality of the set $S_i$ there exists a point $x_i\in S_i$ such that $d(x, x_i)<2\varepsilon$. Therefore $B_{\varepsilon/2}(x_i)\subset B_{5\varepsilon/2}(x)$. From the other hand, the construction of the sets $S_i$ implies that the sets $B_{\varepsilon/2}(x_i)$ and $B_{\varepsilon/2}(x_j)$ are disjoint provided $i\ne j$. Since a ball with radius $5\varepsilon/2$ can cover at most $5^n$ mutually disjoint balls with radius $\varepsilon/2$, $N\le 5^n$.
Remark. For small $n$ better upper bounds can be obtained from circle or sphere packing bounds. For instance, current results (see [Bol] or [Fri]) suggest a bound $DG(\Bbb R^2,d,\varepsilon)\le 20$
Remark. I expect that Proposition 1 can be generalized to any metric $d$ on $\Bbb R^n$ generated by a norm.
Conjecture 1. Let $(A, d)$ be the space $\ell^2$ endowed with the standard metric. Then $DG(A,d,1)=\infty$.
PS. Now I'm going to pose your questions at Monday at our seminar "Topology & Applications", because they are similar to asymptotic (covering) dimension of metric space (especially, of macrofractals), which is investigated by my master Taras Banakh and his teacher Michael Zarichnyi.
References for circle or sphere packing bounds
[Bol] Dave Boll. Optimal Packing Of Circles And Spheres.
[Fri] Erich Friedman. Packing center Circles in Circles.
[Pet] Ivars Peterson. Pennies in a Tray.
