In John Baez's blog post http://www.math.ucr.edu/home/baez/week151.html, Baez gives a short proof of why $S^\infty$ is contractible, using that $S^\infty$ is the unit sphere in a infinite dimensional Hilbert space.
I can't resist explaining why the unit sphere in an infinite-dimensional Hilbert space is contractible. It seems very odd that a sphere could be contractible, but this is one of those funny things about infinite dimensions. Take our Hilbert space to be $L^2[0,1]$ and consider any function f in the unit sphere of this Hilbert space: $$\int|f(x)|^2 dx = 1$$ For $t$ between $0$ and $1$, let $f_t(x)$ be a function that equals $1$ for $x < t$, and a sped-up version of $f$ for $x$ greater than or equal to $t$. If you do this right $f_t$ will still lie in the unit sphere, and you'll have a way of contracting the whole unit sphere down to a single point, namely the constant function $1$. Cute, huh?
This appears to draw a path from each point in $S^\infty$ to $1$. I guess you can form the homotopy that drags each vector along its path to $1$ but I'm not sure how to show this is a contraction. I imagine doing a similar thing for $S^2$, which carries each point along a geodesic to the south pole, but the resulting homotopy would rip apart the sphere at the north pole. Obviously it shouldn't work in $S^2$, but I don't see how this is addressed in infinite dimensions. Am I thinking about this wrong?
Q: How does this proof give a contraction of $S^\infty$?