Let $n\in \mathbb N$ and let $D^n$ be the closed $1$-ball in $\left(\mathbb R^n, \|\,\cdot\,\|_1\right)$. It is not too hard to show that $[0,1]^n \cong D^n$ in this case.

This observation leads to the question whether we also have $[0,1]^\omega \cong D^\omega$, where

$$D^\omega = \left\{(x_n)_{n\in \mathbb N} \in \ell^1 \; \Bigg| \; \sum_{n=1}^\infty |x_n|\le 1\right\}$$

is the closed unit ball in $\ell^1$.

Now if we view $D^\omega$ as a subspace of $\ell^1$, then we can prove that it is not homeomorphic to $[0,1]^\omega$ by arguing that the latter is compact, while the former is not.

But what happens if we view $D^\omega$ as a subspace of $\mathbb R^\omega$?

Is $D^\omega$ homeomorphic to $[0,1]^\omega$, when both sets are endowed with the subspace topologies induced by $\mathbb R^\omega$ (in the product topology)?

In this case both spaces are compact ($D^\omega$ is the intersection over all $n\in \mathbb N$ of the compact sets $\{x\mid \sum_{i=1}^n |x_i| \le 1\} \cap [-1,1]^\omega$) and all topological properties I can think of are preserved when we go from $[-1,1]^\omega \cong [0,1]^\omega$ to the subspace $D^\omega$. I have also tried to explicitly construct a homeomorphism, but to no avail.

I hope some of the topologically savvy guys on this site could help me out here! Thanks =)