TITLE: Finite support iterations of $\sigma$-centered forcing notions
QUESTION [13 upvotes]: I am looking for a proof (or better, a reference) of the following fact: 
The finite support iteration of $\sigma$-centered forcing notions is again $\sigma$-centered, assuming we iterate less than $(2^{\aleph_0})^+$ steps. 
(EDIT: In the first version of the question I forgot to mention that it was  Stefan Geschke who suggested that there should be a proof similar to "the product of continuum many separable spaces is still separable")
EDIT: In his 1994 paper "$\sigma$-centred forcing and reflection of (sub)metrizability" in TAMS (MR1179593 94g:54003), Frank Tall writes: 

"It is well known (proved by the same method that proves the product of $\le 2^{\aleph_0}$ separable spaces is separable) that the finite support iteration of $\le 2^{\aleph_0}$ $\sigma$-centred orders is $\sigma$-centred."

REPLY [7 votes]: I just wanted to point out how Andreas' proof above relates to the separability of products of $2^{\aleph_0}$ separable spaces:
Let $\lambda<(2^{\aleph_0})^+$.
Suppose $\langle X_\alpha:\alpha<\lambda\rangle$ is a family of separable spaces.  For each $\alpha<\lambda$ let $\langle x_\alpha^n:n\in\omega\rangle$ enumerate a dense subset of $X_\alpha$.
Let $F:\lambda\times\omega\to\omega$ be as in Andreas' proof.
Now the set of sequences of the form
$\langle x_{\alpha}^{F(\alpha,n)}:{\alpha<\lambda}\rangle$, $n\in\omega$, is dense in $\prod_{\alpha<\lambda}X_\alpha$.  
Also note that $\sigma$-centeredness of a forcing notion $\mathbb P$ corresponds to the separability of the Stone space of the completion of $\mathbb P$.
However, I don't know whether it is possible to deduce the $\sigma$-centeredness of short finite support iterations of $\sigma$-centered forcing notions directly from the separability of small product of separable spaces.