Connectedness of a metric space implies connectedness of the corresponding Hausdorff hyperspace 
If $(X,d)$ is a connected metric space, show that $(\mathscr H(X),h)$ is also connected.

where h is the Hausdorff distance, define by max$\{d(A,B),d(B,A)\}$
and $d(A,B):=$max$\{$min$\{d(a,b)|a\in A\}|b\in B\}$
$\mathscr H(X)$ is the collection of all compact subset of $X$.


I know what connected means, but it's too hard for me to figure out what is connected actually means in fractal geometry.
 A: Here is a complete proof. It follows the idea in my comment of Hagen von Eitzen's answer. It is more complicated, but I also steamline his idea, so in the end it is not too long.
Let $(X,d)$ be a connected metric space. Let $n \geq 1$. Then $(X^n, d_n)$ is a connected metric space, where $d_n (x, y) = \max_i d (x_i, y_i)$. Let $\pi_n : \ X^n \mapsto \mathcal{H} (X)$ be defined by:
$$\pi_n (x) = \bigcup_{i=1}^n \{x_i\}.$$
Then $\pi_n$ is $1$-Lipschitz, and as such is continuous. Hence, $\pi_n (X^n)$ is a connected space. But $\pi_n (X^n)$ is exactly the set of compact subsets of $X$ of cardinal at most $n$.
Let $\mathcal{U}$ be the connected component which contains $\pi_1 (X^1)$. Then it contain $\pi_n (X^n)$ for all $n$ (since the intersection of $\pi_n (X^n)$ with $\mathcal{U}$ is non-empty, $\pi_n (X^n)$ is connected, and $\mathcal{U}$ is a connected component). Hence, $\mathcal{U}$ contains $\bigcup_n \pi_n (X^n)$. Since $\mathcal{U}$ is closed, it also contains $\overline{\bigcup_n \pi_n (X^n)}$.
But $\overline{\bigcup_n \pi_n (X^n)} = \mathcal{H} (X)$ ! Hence, $\mathcal{H} (X)$ is connected.
NB: Hagen von Eitzen's answer can essentially be written in a much shorter fashion:
"The map $x \mapsto \{x\}$ is an isometry from $X$ to $\mathcal{H} (X)$. Since, $X$ is connected, its image is also connected."
A: Assume $\mathscr H(X)=\mathscr U\cup \mathscr V$ with disjoint $h$-open sets $\mathscr U,\mathscr V$. Let $U=\{\,x\in X\mid \{x\}\in \mathscr U\,\}$, $V=\{\,x\in X\mid \{x\}\in \mathscr V\,\}$. Then $U\cap V=\emptyset$ and (since singletons sets are compact) $X=U\cup V$.
If $x\in U$ then there exists $\epsilon>0$ such that $A\in\mathscr U$ for any compact $A$ with $h(A,\{x\})<\epsilon$. Specifically $\{y\} \in \mathscr U$ whenever $d(y,x)<\epsilon$. We conclude that $U$ is open and by the same argument $V$ is open.
By connectedness, one of $U,V$ must be empty, say $U=\emptyset$.
Remains to show that $\mathscr U$ is empty...
