Let $(X, d)$ be metric space. Define $B_\epsilon = \{ x \in X : \exists b \in B \; d(x, b) \le \epsilon\} $. Let $F(X)$ be a family of all nonempty compact subsets of $X$ (so $\emptyset \notin F(X)$ ). We shall define Hausdorff metric by: \begin{equation} D(A, B) = \inf \{ \epsilon \in \mathbb{R}^+ : A \subset B_\epsilon \; \land B \subset A_\epsilon \}. \end{equation} Then $(F(X), D)$ is a metric space.
What I would like to know is what is the relation between connectedness of $(X, d)$ and connectedness of $(F(X), D)$.
So far I was able to prove that if $(X, d)$ is not connected, then $(F(X), D)$ also is not connected. Here the main idea was that if (A, B) is a pair of nonempty subsets of $X$ such that $A\cup B = X$ and $A \cap B = \emptyset$ and $A, B$ are open, then $(2^A \cap F(X), F(X) \setminus 2^A \cap F(X))$ is a pair of nonempty, open subsets of $F(X)$ which sum to $F(X)$ and have empty intersection. (Which by contraposition means that connectedness of $(F(X), D)$ implies connectedness of $(X, d)$).
Is the opposite implication true, that is, does connectedness of $(X, d)$ imply connectedness of $(F(X), D)$?
$F(X)$ is connected when $X$ is.
Suppose $F(X)$ is disconnected by open subsets $U,V$. That is, they are disjoint collections of compact subsets of $X$ whose union is $F(X)$. Let $A:=\{x\in X:\{x\}\in U\}$, and similarly for $B$.
Let $a\in A$, $b\in B$ be any two points. Then the compact set $\{a,b\}$ is in either $U$ or $V$. Then $D(\{a,b\},\{b\})>\epsilon$ for some $\epsilon>0$, since $U$ and $V$ are open sets. This means that $d(a,b)>\epsilon$. Hence $B(\epsilon,a)\subseteq A$, and $A$ is open. Similarly for $B$, so $X$ is disconnected.