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 $i \neq j$, both in $I$. Then define a (basic) open set $U$ of $X$ by $U= p_i^{-1}[\{0\}] \cap p_j^{-1}[\{1\}]$ (note that we use that $\{0,1\}$ is discrete here!). Then there is some $d \in D \cap U$, as $D$ is dense so intersects all non-empty open subsets of $X$. Then $d_i = 0, d_j = 1$, so $d \in D_j \setminus D_i$, which shows that $D_j$ and $D_i$ are not equal as sets.
