Is the following characterization of a finite set $A$ in a metric space true? Let $(M,d)$ be an infinite metric space with distance function $d$ such that there exists a $B \subseteq M$ bounded infinite set.
Is it true that

$A\subseteq M$ is finite $\stackrel{?}{\leftrightarrow} \exists B\subseteq M$ bounded infinite and $\exists f:A\to B$ injective s.t.
$$\inf_{\begin{matrix}\begin{matrix}a,b\in A \\ a\ne b\end{matrix}\end{matrix}} d(f(a),f(b)) \ne 0$$

$f$ can be thought of as a "packing" function that packs all the values of $A$ into a bounded set $B$. If we can pack the values in a way that even the closest ones are more than distance $0$ apart from each other ($d(f(a),f(b)) > 0$), then the original set $A$ couldn't have contained a "lot" of elements (it should be finite).
The $(\rightarrow)$ direction isn't too difficult to prove:
Let $|A|:=n$, then we can define the diameter of $B$ as the diameter of the largest distance that can fit into it:
$$\text{diam}(B) := \sup_{x,y \in B} d(x,y)$$
For a bounded infinite $B$, this has to be $>0$.
Then lining up the elements of $A$, equally separated by $\frac{\text{diam}(B)}{2n}$ (this is what the $f$ function will do), they will all fit into $B$ and their minimum distance is
$$\inf_{\begin{matrix}a,b\in A \\ a\ne b\end{matrix}} d(f(a),f(b)) = \min_{\begin{matrix}a,b\in A \\ a\ne b\end{matrix}} d(f(a),f(b)) = \frac{\text{diam}(B)}{2n} > 0 \quad \square$$
However, I'm struggling to prove the $(\leftarrow)$ direction. I think intuitively it should be true though. If not, then is there a packing of infinitely many elements where the minimum (infimum) distance between two elements is $>0$?
 A: Aside from the issue mentioned in the comment there is another problem, closely related to the fact that in a metric space closed and bounded sets don't have to be compact. Take for instance the set
$$H=\{a=(a_1,a_2,\dots) \ | \ a_i \in \mathbf{R} \quad \text{and} \quad \sum_{n=1}^\infty a_n^2 < \infty \}$$ with metric
$$d(a,b)=\sqrt{\sum_{n=1}^\infty (a_n-b_n)^2}.$$
You can define a sequence $a^{(1)},a^{(2)},a^{(3)},\dots$ by taking $a^{(m)}$ to be the point with a $1$ in position $m$ and $0$'s elsewhere. The distance between any two distinct points in this sequence is $\sqrt{2}$, and the whole sequence is bounded (all are of distance $1$ from $0$) but it is not finite. So the implication $\Leftarrow$ fails.
A: So first of all I assume that the infimum is taken over $a\neq b$, otherwise it is always trivially $0$.
The "$\Leftarrow$" implication is not true. It is enough to find a bounded infinite subset $B$ such that the infimum over distances of its distinct elements is positive. Take $B=M$ to be any infinite set with the discrete $d(x,y)=1$ metric.
The "$\Rightarrow$" implication is also not true, even when $M$ is assumed to be infinite. Because it doesn't have to have an infinite bounded subset, e.g. $\mathbb{N}$ with the Euclidean distance. But if $M$ has an infinite bounded subset, say $C\subseteq M$ then it is true. You simply take $B:=A\cup C$ and $f$ to be the inclusion $f(x)=x$.
