$A_1, \dots A_n$ nonempty compact in (X, d) can we find $a_1, \dots a_n \in X$ such that $\text{min}\{d(a_i,a_j)\} \geq \text{min}\{d_H(A_i, A_j)\}$ Suppose $(X, d)$ is a compact metric space. Let $K(X, d)$ be a family of nonempty compact subsets of $X$ and let $d_H : K(X, d) \times K(X, d) \rightarrow \mathbb{R}$ be a Haussdorff metric on $K(X, d)$ (https://en.wikipedia.org/wiki/Hausdorff_distance).
I want to prove or disprove this claim:
Suppose we have $A_1, \dots A_n \in K(X, d)$, for some $n \in \mathbb{N}$. Let us define $\rho_n := \text{min}\{d_H(A_i, A_j) \mid i,j \in \{1, \dots, n\}, i \neq j\}.$ Than there exist $a_1, \dots, a_n \in X$ such that $\text{min}\{d(a_i, a_j) \mid i,j \in \{1, \dots, n\}, i \neq j\} \geq \rho_n.$
Since I think this claim is true I tried to prove it.
Firstly if we have only $1$ set $A_1$ it is trivial.
If we have two sets $A_1, A_2$ than by the definition of $d_H$ either there exists $a_1 \in A_1$ such that for every $a_2 \in A_2$ we have $d(a_1, a_2) >= d_H(A_1, A_2) = \rho_2$ or there exists $a_2 \in A_2$ such that for every $a_1 \in A_1$ we have $d(a_1, a_2) >= d_H(A_1, A_2) = \rho_2$. In any case the statement holds. Now I was thinking about using induction but I was not able to prove it like that. I also tried to prove it if $n = 3$, since I think that proof could easily be generalized to arbitrary $n$, but i was not able to do it.
Thank you for any help.
 A: No. Let $X$ be a finite discrete space with $n$ elements. Consider all subsets with two elements. There are $\binom n2$ of them. Then Hausdorff distance between any two of them is 1. However there can are only $n$ distinct elements in $X$.
Edit (after request on the comment):
No, the additional assumption that the space is infinite won't help. Consider the set
$$X=[-1,1]\times\{0\} \cup \{0\}\times [-1,1]$$ equipped with the euclidean topology. Denote $a_\pm=(\pm 1,0)$, $b_\pm=(0,\pm 1)$. Consider six two-element subsets of $\{a_+,a_-,b_+,b_-\}$. The Hausdorff distance between them is equal to $\sqrt 2$. There are at most $4$ points with such distance between each of them.
A: No.
Consider $A_1 =[0,1], A_2 = [2,3], A_3 = [4,5]$.
Then $\min d(A_i,A_j) = 3=d(A_1,A_2)  = d(A_2,A_3)  $
Obviously the $\min d(a_i,a_j)$ will be one of $d(a_1,a_2)$ or $d(a_2,a_3)$.
In order to get $d(a_1,a_2) =3$ we must choose $a_1=0,a_2=3$. But then for any choice of $a_3$ we have $d(a_2,a_3) \le 2$  and so $\min d(a_i,a_j) \le \min d(A_i,A_j)$.
Likewise in order to get $d(a_2,a_3) = 3$ we must have $d(a_1,a_2) \le 2$
