TITLE: Question about 0-dimensional Polish spaces
QUESTION [6 upvotes]: Hello everybody,
I'm stuck with proving (or disproving) the following statement.
Statement:
For every $0$-dimensional Polish space $(X,\mathcal{T}\ )$, and a countable basis of clopen sets $\mathcal{B}$ for $\mathcal{T}$, every open set is the disjoint union of clopen sets in $\mathcal{B}$.
Every open set is the union $O=\cup_{n}B^{0}_{n}$ of the basic clopen sets contained in it (say ordered with a given numbering of $\mathcal{B}$). The idea is to make it a disjoint union by considering, iteratively, 
$O= B^{0}_{1} \cup  O^{1}$
where $O^{1} = O\setminus B^{0}_{1}$, which is open. Then again we have
$O^{1}=\cup_{n}B^{1}_{n}$. 
So consider $O^{2}=  O^{1}\setminus B^{1}_{1}$
etcetera. The resulting union 
$\cup_{m} B^{m}_{1}$  
is open. However, I'm stuck in proving that, in general, a point $x\in O$ ends up necessarily in some $B^{k}_{1}$, for $k\in \mathbb{N}$, i.e., I can't prove that 
$O= \cup_{m}B^{m}_{1}$.
Googling around I found this interesting paper [1]. The author says that it is a known fact (unfortunately he doesn't give a reference) that for every $0$-dimensional Polish space, the Borel sets are generated from the clopens by closing under countable disjoint unions and complements. This does not solve my problem, but still, I would  be interested in reading a proof. Could you point me to some relevant literature? 
Thanks in advance,
[1] Abhijit Dasgupta. Constructing $\Delta^{0}_{3}$ using  topologically restrictive countable disjoint unions.

REPLY [2 votes]: If you have such a union of $O = \cup_n B_n$, consider the sets $B'_n = B_n \setminus \cup_{i=0}^{n-1} B_i$, where $B'_0 = B_0$. Show these sets are disjoint, clopen, and have the same union as the original sets, as for every $x \in O$ there is a first index $n(x)$ such that $x \in B_{n(x)}$.