Why can any open subset $U$ of $\mathbb{Q}^\infty$ be written as disjoint union of basic clopen subsets? Edit: I asked the same question on mathoverflow, but I don´t completely understand the answer I got (even if I believe it is great.) Therefore I rephrased the answer here and will appreciate if someone posts another proof or helps explain the proof here).
I am reading Engelen´s paper and have trouble with this proof of Lemma 2.1 (a) (link is below).

It is easily seen that any non-empty open subspace $U$ of $\mathbb{Q}^\infty$ can be
written as an infinite disjoint union of non-empty basic clopen
subsets; hence, $U = \mathbb{N} \times \mathbb{Q}^\infty \simeq \mathbb{Q}^\infty$.

In particular, I don't see how the $U$ can be written as the said disjoint union and how that implies that $U \simeq \mathbb{Q}^\infty$
Can anyone please help with this or provide any reading source? I would be really grateful.
Definitions:
$\mathbb{Q}^\infty$ is defined as a set of all rational sequences, endowed with the standard product topology.
Source: Engelen - Countable Product of zero-dimensional absolute $F_{\sigma \delta}$ spaces, Lemma 2.1 (a).
 A: My answer in MO was written hastily, here is the same proof but in more detail (and better notation):
First, for each sequence $s=(S_n)_{n\in\mathbb{N}}$ of clopen sets of $\mathbb{Q}$, let $A_s$ be the subset $\prod_{n\in\mathbb{N}} S_n$ of $\mathbb{Q}^\mathbb{N}$. Note that if $S_n=\mathbb{Q}$ for all $n$ except finitely many, then $A_s$ is clopen. I will call sets of this form "basic clopens". As $\mathbb{Q}$ has a basis formed by clopen sets (e.g. intervals with irrational ends), basic clopens form a basis for the topology of $\mathbb{Q}^\mathbb{N}$.
Also note that nonempty basic clopens are homeomorphic to $\mathbb{Q}^\mathbb{N}$. This is a consequence of the fact that any nonempty clopen of $\mathbb{Q}$ is homeomorphic to $\mathbb{Q}$ by Sierpinski's Theorem.
For any two basic clopens $A_s$ and $A_t$, the clopen $A_s\cap A_t$ is given by $A_r$, where $r=(R_n)_n$ is given by $R_n=S_n\cap T_n$. So it is also a basic clopen.
Lemma: The complement of any basic clopen $A_s$ is a finite union of basic clopens.
Proof: For each sequence $\xi$ of elements of $\{-1,1\}$, consider the clopen $A_{s_\xi}$, where $s_\xi$ is the sequence given by $S_n$ if $\xi_n=1$ or $\mathbb{Q}\setminus S_n$ if $\xi_n=-1$. Note that $A_{s,\xi}$ is empty for all $\xi$ except finitely many, and if for some $\xi$ we have that $A_{s,\xi}$ is nonempty, then it is clopen. Moreover, the sets $\{A_{s_\xi}\}_{\xi\in\{-1,1\}^\mathbb{N}}$ are pairwise disjoint and their union is all $\mathbb{Q}^\mathbb{N}$. So the complement of $A_s=A_{s_{(1,1,1,\dots)}}$ is just the union of the rest of clopens $A_{s,\xi}$, with $\xi\neq(1,1,1,\dots)$, which are nonempty. $\square$
Corollary: If $A_1,\dots,A_n$ are basic clopens, then $A_n\setminus\bigcup_{i=1}^{n-1}A_i$ is a finite union of basic clopens.
Proof: $A_n\setminus\bigcup_{i=1}^{n-1}A_i=\bigcap_{i=1}^{n-1}(A_n\setminus A_i)$. Each set $A_n-A_i$ is a finite union of basic clopens, because it is $A_n\cap(\mathbb{Q}^\mathbb{N}\setminus A_i)$. So their intersection is also a finite union of basic clopens.$\square$
Now consider any open subset $U$ of $\mathbb{Q}^\mathbb{N}$. As $\mathbb{Q}^\mathbb{N}$ is second countable and basic clopens form a basis for its topology, we can express $U=\cup_{n\in\mathbb{N}}U_n$, where the $U_n$ are all basic clopens. Letting $V_n=U_n\setminus\bigcup_{i=1}^{n-1}U_i$, we also have $U=\cup_n V_n$. The $V_n$ are pairwise disjoint, and each one is a finite union of disjoint clopens. So we have obtained a form to express $U$ as a union of disjoint clopens.
We also want infinitely many of the clopens to be non-empty: for that, it is enough to have infinitely many non empty $V_n$: this can be achieved by choosing the cover $U_n$ of $U$ in such a way that no finite union of the $U_n$ cover $U$.
To obtain a cover of $U$ by clopens such that no finitely many of them cover $U$, we can do the following: let $\pi_m:\mathbb{Q}^\mathbb{N}\to\mathbb{Q}$ be the m$^\text{th}$ projection of the product $\mathbb{Q}^\mathbb{N}$ onto its factors. As $U$ is open, we have $\pi_m(U)=\mathbb{Q}$ for some $m\in\mathbb{N}$. Now take an arbitrary open cover $U_n$ of $U$ and let $U_{n,k}=U_n\cap\pi_m^{-1}((-k\pi,k\pi))$. The cover $(U_{n,k})_{n,k}$ of $U$ satisfies what we want because for any union $X$ of finitely many of the $U_{n,k}$, $\pi_m(X)$ is bounded, so $X\neq U$.
So, for any open set $U$ of $\mathbb{Q}^\mathbb{N}$, we have obtained $U$ as an infinite union of disjoint basic clopens, which in turn are homeomorphic to $\mathbb{Q}^\mathbb{N}$. So, $U$ is homeomorphic to $\mathbb{N}\times\mathbb{Q}^\mathbb{N}$. Note that this also aplies to the open set $U'=\mathbb{Q}^\mathbb{N}$. So $\mathbb{Q}^\mathbb{N}$ is also homeomorphic to $\mathbb{N}\times\mathbb{Q}^\mathbb{N}$, thus $U$ is homeomorphic to $\mathbb{Q}^\mathbb{N}$.
Note: A similar, easier proof works to show that any open subset of the Cantor set $\mathcal{C}\subseteq[0,1]$ which is not compact is homeomorphic to $\mathbb{N}\times\mathcal{C}$: we can take a base of clopen sets of the Cantor set (which are homeomorphic to $\mathcal{C}$ by Brouwer's theorem). Then we can express any open subset $U$ of $\mathcal{C}$ as a union of a sequence $A_n$ of these clopen sets. Letting $B_n=A_n\setminus\bigcup_{i=1}^{n-1}A_i$, $B_n$ are clopen so they are either empty or homeomorphic to $\mathcal{C}$. Moreover, as $U=\bigcup_nB_n$ is not compact, infinitely many of the $B_n$ have to be non empty. So $U\simeq\mathbb{N}\times\mathcal{C}$.
A: This is my own attempt to answer this question. I am using answer by Saúl RM on stackoverflow, where I posted the same question, so credits go to them. Please correct me if I am wrong in anything, thank you.

(1) Goal of the proof is to see that $U \simeq \mathbb{Q}^\omega$ where $U$ is nonempty and open.
First, observe that we can express any such $U$ as countable union of basic clopen subsets $(A_n)$ where for each $n \in \mathbb{N}$, $A_n =\{(q_k)_{k\in\mathbb{N}};i_{n,k}<q_k<j_{n,k} \text{ for $k=1,\dots,n$}\}$, where $i_{n,k}$ and $j_{n,k}$ are irrational numbers.
Each such $A_n$ is therefore a sequence of rational numbers $q_k$ located between pairs of irrational numbers $i_{n,k}, j_{n,k}$, such that for the fixed $n$, we have one pair of such numbers for each $k \leq n$. In total, this will give us $n$ pairs of irrational numbers and hence $n$ sequences of rationals (one between each pair).
Important is that some $A_n$ can be empty, only finitely many $A_n$ may be nonempty. $U$ can be always written as their union, but we can have a situation like $U = A_1$.
(2) Now, we need to express $A_n\setminus\bigcup_{i=1}^{n-1}A_i$ as a finite union of disjoint basic clopens. Here, I don´t really understand why we can do that.
(3) To do the abovesaid, we can work in $\mathbb{Q}^n$ instead of $\mathbb{Q}^\omega$ due to the construction of the sets $A_n$. This I also don´t really understand.
(4) Writing $\mathbb{Q}^n=\mathbb{Q}_1\times\dots\times\mathbb{Q}_n$, notice that for each $k\leq n$, the irrationals $i_{1,k},\dots,i_{n,k},j_{1,k},\dots,j_{n,k}$ give a partition of $\mathbb{Q}_k$ into finitely many clopen subsets. Taking a product of these partitions, we obtain a partition of $\mathbb{Q}^n$ into disjoint clopens such that any of the $A_i$, with $i\leq n$, is a finite union of these clopens. So $A_n\setminus\bigcup_{i=1}^{n-1}A_i$ is also a finite union of the clopens.
(5) If we pick the $A_n$ so that no finite union of them covers $U$, we also get infinitely many non empty clopens. I also don´t understand this step.
(6)
This implies that $U\equiv\mathbb{Q}^\infty$ because any basic clopen set like the ones above is homeomorphic to $\mathbb{Q}^\infty$, which can be deduced using that any clopen subset of $\mathbb{Q}$ is homeomorphic to $\mathbb{Q}$ by Sierpinski's theorem.
