Let $m\in \mathbb{N}^*$ be a fixed number, and set $$S_m:=\{y\in \mathbb{C}^m; \|y\|=1\}$$
Consider the $\ell^2$ complex Hilbert space. I want to show that $S_m\times \ell^2$ cannot be homeomorphic to $$ S(0,1)=\left\{ x=(x_n)_n\subset \ell^2\ :\ \sum_{n=1}^\infty |x_n|^2=1\right\}.$$
To provide the carrying out of deyore's comment:
As a map $S(0,1)\to S(0,1)$ the left shift $$\sum_k x_k e_k\mapsto \sum_k x_k e_k$$ is homotopic to identity. If you compose this homotopy with: $$h_t\left(\sum_kx_k e_{k+1}\right)= \sqrt{t}\, e_1+\sqrt{1-t}\sum_k x_k e_{k+1}$$ you will contract to the point $e_1$.
The problem is to show this left shift is homotopic to identity, here you need to see that for all $t$ the map $(1-t)\Bbb1 +tL$ has zero kernel, where $L$ is the left shift. Then the map $x\mapsto \frac{(1-t)x+t Lx}{\|(1-t)x+tLx\|}$ is a homotopy on $S(0,1)$ between the identity and left shift.
Specifically suppose you have a $t$ and an $\sum_k x_ke_k\in \ell^2$ so that
$$[(1-t)\Bbb1+tL]\sum_k x_k e_k =0= \sum_k [(1-t)x_k+tx_{k-1}]e_k$$ Where we define $x_{-1}=0$. Assume $t\neq1$, then the $k=0$ part implies $x_0=0$, plug this into $k=1$ to retrieve $x_1=0$ and so on. Thus the kernel of this map is $0$.