I know that the unit ball in $\mathbb{R}^n$ can be a countable union of rectangles $[a_1,b_1)\times\cdots\times[a_n,b_n)$.
The question I have is whether the unit ball in infinite dimensional space, e.g $\ell^2$, can also be a countable union of rectangles $\prod_{n=1}^{\infty}[a_n,b_n) := \lbrace (x_n)_{n\in\mathbb{N}}\mid x_n\in[a_n,b_n),\forall n\in\mathbb{N}\rbrace$.
In the finite dimensional case:
$B_1(0) = \bigcup_{q_k\in\mathbb{Q}_+,q_1^2+\dots+q_n^2<r^2}[-q_1,q_1)\times\dots\times[-q_n,q_n)$
But this does not work in the case of $\ell^2$, since $\mathbb{Q}^{\mathbb{N}}$ is uncountable.
I am starting to think that the result is false in the infinite dimensional case, but I currently have no idea.
The first thing that should come to mind is to try a Cantor diagonalization argument, which indeed seems to work.
Suppose that in $\ell^2$, we can write $B_1(0)=\bigcup_{n\in\mathbb{Z}_{>0}}P_n$, where for all positive integers $n$ the set $P_n$ is a product of intervals, i.e. $P_n=\prod_{k=1}^{\infty}I_{n,k}$, where $I_{n,k}$ is an interval. Denote $$ \mu_{n,k}:=\sup_{x\in I_{n,k}}|x| $$ and notice that as $P_n\subseteq B_1(0)$, we must have $\sum_{k>0}\mu_{n,k}^2\leq 1$. Therefore, for all $n>0$, there exists $K_n>0$ such that $\mu_{n,k}<2^{-n}$ for all $k\geq K_n$. Therefore, we can inductively construct an increasing sequence $k_1<k_2<\cdots<k_n<\cdots$ such that $\mu_{n,k_n}<2^{-n}$. In particular, we have $2^{-n}\notin I_{n,k_n}$. Hence, the sequence $\underline{x}=(x_k)_{k\in\mathbb{Z}_{>0}}$ defined by $$ x_k= \begin{cases} 2^{-n} & \text{if }k=k_n\text{ for some }n>0,\\ 0 & \text{otherwise.} \end{cases} $$ is not contained in any $P_n$. On the other hand, we have $\|\underline{x}\|_{\ell^2}=\sum_{n>0}2^{-2n}=1/3<1$ and thus $\underline{x}\in B_1(0)$, contradiction. Hence $B_1(0)$ cannot be written as the countable union of rectangular boxes.