**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).