Is the hyperspace of the Hilbert cube homeomorphic to the Hilbert cube Question: Is the hyperspace of the Hilbert cube $H=[0,1]^\mathbb {N}$ homeomorphic to $H$?
Remarks and definitions: 
1) The Hilbert cube $H$ is a compact metric space, where the metric is given by the $\ell_2$-norm of sequences. A classical theorem on metric spaces says that every compact metric space is isometric to a closed subspace of $H$. 
2) The hyperspace of a metric space $X$ is the metric space of all non-empty compact subsets of $X$ given by the Hausdorff metric. Another classical theorem on metric spaces says that the hyperspace of a compact metric space is again a compact metric space.
Combining 1) and 2) shows that the hyperspace of the Hilbert cube is isometric to a closed subspace of the Hilbert cube. So my question asks whether we also can get a homeomorphism 
(can we even get both spaces isometric?).
 A: The hyperspace of any Peano continuum (locally connected metric continuum)  is homeomorphic to the Hilbert cube; this is a result of Curtis and Schori, see here.
I learnt about this result in a paper of Torunczyk, where a different proof is given (that paper is also available online)
When it comes to asking about an isometry, I'm not sure which metric you put on the Hilbert cube...
A: Rule out isometry ... In the hyperspace of $\mathbb R$, let $A=\{0,1,2\}$, $B=\{0\}$, $C=\{2\}$ and $M = \{1\}$.  Then points $A, B, C$ all have distance $2$ from each other; they are the vertices of an equilateral triangle.  But $M$ has distance $1$ from each of $A, B, C$, so $M$ is the midpoint of each of the three sides of that triangle.  Not possible in Euclidean geometry.
A: Part 1) of ''Remarks and definitions'' is incorrect:
I truly do not understand what you meant when you wrote ''the metric is given by the $\ell_2$-norm of sequences''.
Nevertheless, whatever the metric is on $H$, it is not true that every compact metric space embeds into $H$ isometrically. As a compact space, it is bounded, hence there is a two-point metric space that cannot be embedded into $H$ isometrically.
Also it is a nice exercise two show that if $X$ is a compact metric space, then $X$ is not isometric to any of its proper subspaces $Y\subsetneqq X$.
