arbitrary product of separable spaces not always is separable I have the following question:
Let $X_i = (\{0,1\}, \operatorname{discrete})$ where $I= 2^{\mathbb{R}}$ the set of index and let $$X= \{0,1\}^{2^{\mathbb{R}}} = \prod_{i \in I}X_i$$
let $D$ a dense set such that $D \subseteq X$ and  let the set
$$D_i := D \cap p_i^{-1}(1)$$. Prove that for each $i \neq j$ we have that $D_i \neq D_j$
I have tried the following:
Suppose that is true that $D_i = D_j$, then we have that $D \cap p_i^{-1}(1) = D \cap p_j^{-1}(1)$, so inmediatly we have that $ p_i^{-1}(1)= p_j^{-1}(1)$ but this is not true since the projection just will coincide in one point at most.
am I in the correct way?, I am really confused because I don´t know how can I express this more formally
I really appreciate your help

Let me offer a completely different strategy, one which shows the beauty of non-constructive proofs. This is a proof in three steps, none of which is particularly difficult.


*

*The product of Hausdorff spaces is Hausdorff. Given two points in the product, pick one coordinates where they differ, and construct basic open sets separating these points using the sets which separate the projections onto the chosen coordinate.

*If $X$ is a separable Hausdorff space, then $|X|\leq 2^{2^{\aleph_0}}$. If $X$ is separable then there is a countable set $D$ whose limit points are everything. Note that the function which maps $x$ to the set $\{A\subseteq D\mid x\in\overline D\}$ is an injective function (due to the Hausdorff-ness of $X$); and its range is included in $\mathcal{P(P(}D))$.

*$2^{2^\Bbb R}$ is too big.
