cover a bounded set with numerable balls (sufficient conditions) metric spaces This natural property is not true always, when I mean a bounded set, I mean a set that it's contained in some ball of radius $r$. It's not always true that if this is the case, I can cover fixing a radius $u < r$  cover with balls of radius $u$, my bounded set, using only finite or even numerable balls. An example is the discrete metric, and any no numerable set ($[0,1]$) . What conditions I can impose in my space, to have this nice property? In any bounded set, not only in compact sets (or at least in open balls, that's what I need now)
 A: If Countable Choice, then your property is equivalent to being a Lindelöf space.
Proof:

I assume we must have $0<u$.  $ \;  $  (I am not sure if that is part of the definition of balls.)
Let $B_u(x)$ be the open ball of radius $u$ centered at $x$.
Let $\overline{B}_u(x)$ be the closed ball of radius $u$ centered at $x$.
Clearly, $\; B_u(x) \subseteq \overline{B}_u(x) \subseteq B_{\frac12 \cdot (u+r)}(x)$,

so it does not matter whether we cover with open or closed balls.
I also assume you are including finite sets as "numerable", and I include them as "countable".

part 1: Lindelöf spaces have your property

Let $C$ be the set of all open balls of radius $u$.

Since the space is Lindelöf, let $C'$ be a countable subcover of $C$.

$C'$ is a countable cover of every bounded subset, and this works for all $u$, which proves part 1.

part 2: If countable choice, then spaces with your property are Lindelöf

(CC) will mean a use of countable choice.

If the space is finite, then in space is compact, in which case the space is Lindelöf.

Assume the space is infinte.  $ \;  $  Let $x$ be a point in the space.

By your property, for all non-negative integers $n$,

there is a countable cover of $B_{n+1}(x)$ by balls of radius $\frac1{n+1}$.

(CC) $\;$ For all non-negative integers $n$, let $C_n$ be such a cover.

For all positive integers $n$, since $x_0\in B_n(x)$, $C_n$ is non-empty.
Define $\quad D \; = \; \displaystyle\bigcup_{n\in \{0,1,2,3,...\}} C_n$
(CC) $\;$ By the last property proven here, $D$ is countable.

Since the diameter of the balls in $C_n$ goes to zero as $n$ goes to infinity, and the space is infinite,

the number of balls in $C_n$ must go to infinity as $n$ goes to infinity.  $ \;  $This shows that $D$ is infinite.  $\;\;$  Since $D$ is countably infinite, let $\; f : \{0,1,2,3,...\} \to D \;$ be some bijection.

(CC) $\;$ For all non-negative integers $n$, let $x_n$ be a member of $f(n)$.

That finishes the setup.

Now, let $U$ be a non-empty open subset of the space, and let $y$ be a member of $U$.

Let $r$ be a positive real number such that $\; B_r(y) \subseteq U \;$.

Since the diameter of the balls in $C_n$ goes to zero as $n$ goes to infinity, there is an $n$ such that the diameter of the balls in $C_n$ is less than $r$.  $\;\;$  Let $m$ be such an $n$.

Let $B$ be a member of $C_m$ such that $y\in B$.  $ \;  $  By construction, $\; x_{f^{-1}(B)} \in B \subseteq B_r(y) \subseteq U \;$.

This works for all non-empty open subsets $U$ of the space, so the space is separable.

(CC) $\;$ Therefore, by Theorem 3.12, the space is Lindelof.

QED
