We can first express the open set $U$ as a countable union of basic clopen subsets $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. Some $A_n$ may be empty.
To answer the question it is enough to express $A_n\setminus\bigcup_{i=1}^{n-1}A_i$ as a finite union of disjoint basic clopens. To do it, we can work in $\mathbb{Q}^n$ instead of $\mathbb{Q}^\infty$ due to the construction of the sets $A_n$.
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.
If we pick the $A_n$ so that no finite union of them covers $U$, we also get infinitely many non empty clopens.
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.