Why is the Hilbert Cube homogeneous?

In Jan van Mill's books (infinite dimensional topology, a prerequisite; the infinite-dimensional topology of function spaces) there are complete self-contained proofs (starting from basic stuff about separable metric spaces) [in essence the same proof in both, but the second book is more recent so might be more readily available]. It is based on Anderson's proof of this, and uses the Inductive Convergence criterion, which states in essence that when (for compact metrisable $X$) $h_n$ is a sequence of homeomorphisms of $X$, there is a sequence of $\epsilon_n > 0$ such that if $d(h_n, 1_X) < \epsilon_n$, where $1_X$ is the identity on $X$, and $d$ is a complete distance on the space of homeomorphisms of $X$, that then the limit of $h_n \circ h_{n-1} \circ \ldots h_1$ as $n$ tends to infinity exists and is a homeomorphism of $X$. Also, each $\epsilon_n$ depends only on the $h_i$ with $i < n$. This allows one to construct homeomorphisms of compact metrisable spaces in a very controlled way. Note that a metric on the Hilbert cube is such that a movement in a high coordinate only contributes a very small amount to the distance, so we can push problems to higher and higher coordinates and "solve" them there by a sufficiently small movement. With this criterion the proof is just 2 pages or so.

When I first saw the question, I intended to post the following link as an answer, but then I saw that Qiaochu had come first with an essentially equivalent answer. I still don't see anything lacking in his answer (I was the first to upvote it), but if someone is looking for a different reference, here is one:

http://www.narcis.info/publication/RecordID/oai:cwi.nl:7603