Here I mean the version with all but finitely many components zero.
This is the swindle, isn't it?

There's an elegant way to phrase this with lots of sines and cosines, but working it all out is too much like hard work.  Here's the quick and dirty way.

Let $T: S^\infty \to S^\infty$ be the "shift everything down by 1" map.

Then for any point $x \in S^\infty$, $T(x)$ is not a multiple of $x$ and so the line between them does not go through the origin.  We can therefore define a homotopy from the identity on $S^\infty$ to $T$ by taking the homotopy $t x + (1 - t)T(x)$ and renormalising so that it is always on the sphere (incidentally, although you are working in $\ell^0$, by talking about a sphere you implicitly have a norm).

Then we simply contract the image of $T$, which is a codimension 1 sphere, to a point not on it, say $(1,0,0,0,0,...)$.  Again, we can use 'orrible sines and cosines, but renormalising the direct path will do.

(Incidentally, there's nothing special about which space you are taking the sphere in.  So long as your space is stable in the sense that $X \oplus \mathbb{R} \cong X$ then this works)

Added a bit later: Incidentally, if you want to work in a space that doesn't support a norm (such as an infinite product of copies of $\mathbb{R}$) you can still define the sphere as the quotient of $X$ without the origin by the action of $\mathbb{R}^+$.  The argument above still works in this case.

Added even later: Revisiting this in the light of the duplicate: Is $L^p(\mathbb{R})$ minus the zero function contractible?, the key property on $T$ is that it be continuous, injective, have no eigenvectors, and be not surjective.  These conditions imply the following:


injective ⟹ the end-point of the homotopy is not the origin
no eigenvalues ⟹ the homotopy does not pass through the origin en route
not surjective ⟹ there is a point not in the image to which the image can be contracted
continuous ⟹ the homotopy is jointly continuous


Finally, there's no difference between the sphere and the space minus a point (indeed, without a norm the "space minus a point" is easier to deal with).  Indeed, the homotopy described here actually works on the "space minus a point" and is just renormalised to work on the sphere.
