A cylinder in infinite-dimensional Hilbert space cannot be homeomorphic to a sphere

The sphere $S(0, 1)$ in the Hilbert space is contractible (Reference). I include the proof given by Loop Space for convenience:

1. $S(0, 1)$ is mapped onto its own equator $S(0, 1)\cap \{x_1=0\}$ by the forward shift $T$.
2. The map $T$ is homotopic to the identity map via the normalized straight-line homotopy $(tx + (1-t)Tx)/\|tx + (1-t)Tx\|$, $0\le t\le 1$.
3. The map $T$ is also homotopic to the constant map $Cx = (1, 0, 0, \dots)$, again via the normalized straight-line homotopy.

On the other hand, the set $S$ is homotopy equivalent to finite-dimensional sphere $S^m$ via the maps $$f:S\to S^m, \quad f(x) = (x_1/1, x_2/2, x_3/3, \dots, x_m/m)$$ $$g:S^m\to S, \quad g(x) = (1 x_1, 2x_2, 3x_3, \dots, mx_m, 0, 0, \dots)$$ Indeed, both $f$ and $g$ are continuous, $f\circ g$ is the identity on $S^m$, and $g\circ f$ is the projection $x\mapsto (x_1, \dots, x_m, 0, 0, \dots)$ which is homotopic to the identity via straight-line homotopy.

And a finite dimensional sphere is not contractible.