How do you show that $S^{\infty}$ is contractible?

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:

1. injective ⟹ the end-point of the homotopy is not the origin
2. no eigenvalues ⟹ the homotopy does not pass through the origin en route
3. not surjective ⟹ there is a point not in the image to which the image can be contracted
4. 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.

Kind of late to the party, but the (weak) contractibility follows from $\pi_i(S^\infty) = 0$ for $i>0$.

General fact: Let $A_{-1} \subset \ldots A_n \subset \ldots$ be a filtration of cellular inclusions of $CW$ complexes. (More generally, let this be a filtration of cofibrations). Then $A_n$ contractible in $A_{n+1}$ $\implies$ $A:=\operatorname{colim}_n A_n$ is contractible. (Here $A$ is given the weak topology.)

Proof: Consider the composition $A_n \times I \xrightarrow{\text{contraction}} A_{n+1} \to A$. Since $A_n \to A$ is a cofibration, extend the above map to a map $A \times I \xrightarrow{\alpha_n} A$. The map $f: A\times I \to A$ defined by $f|_t=\alpha_{n+1}(2^{n+1} t-2^{n+1}+2)\circ\alpha_n(1)\circ\ldots\circ\alpha_1(1)$ for $1-\frac{1}{2^n}\leq t \leq 1-\frac{1}{2^{n+1}}$, is the required retraction. It is continuous because $f|_t$ is continuous when restricted to each $A_n$ and obviously $f|_a$ is continuous for all $a \in A$.

Now give $S^\infty$ the canonical $\mathbb{Z}/2$ equivariant cell structure (i.e. the pullback of the canonical cell structure on $RP^\infty$). The skeletal filtration satisfies the hypotheses of this general fact: $S^n \xrightarrow{i} S^{n+1}$ is null homotopic: $S^{n+1}$ can be given an $n$-skeleton that is a point. By the cellular approximation theorem, the map $i$ is homotopic to one that factors through this particular $n$-skeleton.

I guess this is more complicated than the other answers but this shows that a lot of other things are contractible too (like Milnor space).