I know that the question: ''Is $S^{\infty}$ contractible?'' has already been asked, but none of those post seem to answer it the way I need (Maybe they do, and my lack of knowledge prevents me from aknowledging it)
So, let me define $S^{\infty}$ as the C-W complex such that its n-skeleton is $S^n$ for every $n$.
Now what I thought is this:
$S^2$ is simply connected, then naming $N=(0,0,1)$ you have that $S^1 \subset S^2$ is homotopic to $N$.
(i.e. There exist an homotopy $f_t: S^2\times \mathbb{R} \to S^3$ such that $f_0 \equiv id$ and $f_1(S^1)=\{N\}$)
So, defining $F_0: S^{\infty} \to S^{\infty}, F_0(x)=x \, \, \, \forall x \in S^{\infty}$. i.e. the identity on $S^{\infty}$.
You have that $S^1$ is a subcomplex of $S^{\infty}$, and thus $(S^{\infty},S^1)$ has homotopy extension propperty.
So you have $F_0$, $f_t: S^1 \to S^{\infty}$ (This is an abuse of notation for $f_t$ since the one defined above has co-domain $S^3 \subset S^{\infty}$, not $S^{\infty}$). You also have that $F_0\restriction_{S^1}=f_0$, so by homotopy extension propperty you can extend $f_t$ to an homotopy $F_t: S^{\infty}\to S^{\infty}$, such that $F_t\restriction_{S^1}=f_t$ and $F_0=id$.
So, my problem is that $F_1$ is not necesarilly a constant function, it is constant in $S^1$ but outside of it, I do not know. The other thing I thought was, doing the exact same reasoning with $G_0$ the constant function of value $N$, instead of $F_0$ and finding an homotopy $G_t$, and then try to prove $G_1$ and $F_1$ are homotopic. But I get nowhere, so if anyone has a tip or a suggestion it would be greatly appreciated. Thanks in advanced.