Connectedness of the Hausdorff distance. Does anyone know a proof of connectedness of the Hausdorff distance?
I mean a proof of the following:
Theorem
If $(X, \rho )$ is a connected metric space, then $(F(X), d_h )$ is also connected.
Remark: 
$F(X) = \{ A \subseteq X | A \text{ is nonempty, closed and bounded } \} $, and
$d_h$ stands for the Hausdorff distance.
 A: As stated, the claim is wrong: Consider $X=\mathbb R$ with $\rho(x,y)=\max\{|x-y|,1\}$. Then $X$ has the same topology as standard-$\mathbb R$, espeically it is connected - but all subsets are $\rho$-bounded.
The Hausdorff distance between a standard-bounded and a standard-unbounded subset is at least $1$ so that the sets of standard-bounded closed subsets and the set of standard-unbounded closed subsets are open and witness that $F(X)$ is not connected.
On the other hand, if we change the definition of $F$ to mean
$$F(X)=\{\,A\subseteq X\mid A\text{ nonempty and compact}\,\} $$
the claim is true:
Let $U\subseteq F(X)$ be open. 
If $A\in U$ then there exists $\epsilon>0$ such that $A'\in U$ for all compact sets withd $d_h(A,A')<\epsilon$.
From the open cover $\bigcup_{x\in A}B_{\epsilon/2}(x)$ we pick a finite subcover $\bigcup_{i=1}^nB_{\epsilon/2}(x_i)$. Then $A':=\{x_1,\ldots,x_n\}$ is compact and $d_h(A,A')<\epsilon$, hence $A'\in U$.
Thus

Every nonempty open subset of $F(X)$ contains a finite set.

For finite (possibly empty) $T\subseteq X$ and any $U\subseteq F(X)$ let 
$$S(T,U)=\{\,x\in X\mid T\cup\{x\}\in U\,\}. $$
As $d_h(T\cup\{x\},T\cup\{y\})\le\rho(x,y)$, we see that $S(T,U)$ is open if $U$ is open and closed if $U$ is closed.
Thus if $U\subseteq F(X)$ is clopen then by connectedness of $X$ either $S(T,U)=\emptyset$ or $S(T,U)=X$.
Assume $U\subseteq F(X)$ is clopen and nonempty. Then $U$ contains finite sets. Among these let $A$ be of minimal cardinality. Pick $a\in A$ and let $T=A\setminus\{a\}$. Then $S(T,U)=X$ because $a\in S(T,U)$. If $T\ne\emptyset$ this implies $T\in U$, contradicting minimality of $A$. Therefore $T=\emptyset$, i.e., $U$ contains all singletons.
Consequently, nonempty clopen subsets of $F(X)$ cannot be disjoint and so $F(X)$ is connected.
