Not sure if this is part of what you're wondering about, but will fill in the proof Henno omitted (slightly too long for a comment).

Let $\kappa >|\mathbb R|,$ $U$ be some open proper subset of $I=[0,1],$ and for $\alpha<\beta<\kappa$ define $U_{\alpha,\beta} \subseteq I^\kappa$ to be the basis open set with $U$ at the $\alpha$-th, position, $I\setminus U$ at the $\beta$-th position and $I$ everywhere else. Let $D\subset I^\kappa$ be countable and label $D=\{f_1,f_2,\ldots\}.$

Then, for $\alpha<\kappa$ define the subset of $\mathbb N$ $$ A_\alpha = \{i\in\mathbb N: f_i(\alpha)\in U\}.$$ Since $\kappa > 2^{\mathbb N},$ by pigeonhole, there are $\alpha<\beta < \kappa $ such that $A_{\alpha}=A_\beta.$ So $\forall f\in D,$ either $f(\alpha)\in U$ and $f(\beta)\in U$ or $f(\alpha)\in I\setminus U$ and $f(\beta)\in I\setminus U.$ Thus $D\cap U_{\alpha,\beta} = \emptyset,$ so $D$ is not dense.

Not sure if this is part of what you're wondering about, but will fill in the proof Henno omitted (slightly too long for a comment).

Let $\kappa >|\mathbb R|,$ $U$ and $U’$ be disjoint, open proper subsets of $I=[0,1],$ and for $\alpha<\beta<\kappa$ define $U_{\alpha,\beta} \subseteq I^\kappa$ to be the basis open set with $U$ at the $\alpha$-th, position, $U’$ at the $\beta$-th position and $I$ everywhere else. Let $D\subset I^\kappa$ be countable and label $D=\{f_1,f_2,\ldots\}.$

Then, for $\alpha<\kappa$ define the subset of $\mathbb N$ $$ A_\alpha = \{i\in\mathbb N: f_i(\alpha)\in U\}.$$ Since $\kappa > 2^{\mathbb N},$ by pigeonhole, there are $\alpha<\beta < \kappa $ such that $A_{\alpha}=A_\beta.$ So $\forall f\in D,$ either $f(\alpha)\in U$ and $f(\beta)\in U$ or $f(\alpha)\in I\setminus U$ and $f(\beta)\in I\setminus U.$ Thus $D\cap U_{\alpha,\beta} = \emptyset,$ so $D$ is not dense.