Consider $(0,1)$ with the usual topology, and let $\mathcal{V}$ be a family of open sets. Say that $\mathcal{V}$ is nested iff $V \subseteq V'$ or $V' \subseteq V$ for any $V,V' \in \mathcal{V}$. Call a family $\mathcal{U}$ of open sets ($\subseteq$-)order-dense in $\mathcal{V}$ iff for any $V \subsetneq V'$ in $\mathcal{V}$, there is some $U \in \mathcal{U}$ such that $V \subsetneq U \subsetneq V'$.
(Q) Is there, for any nested family $\mathcal{V}$ of open sets, a countable family $\mathcal{U}$ of open sets that is order-dense in $\mathcal{V}$?
Motivation: by e.g. this answer, the answer to (Q) is 'yes' if and only if any nested family $\mathcal{V}$, equipped with the order $\subseteq$, is order-isomorphic to a subset of $[0,1]$ equipped with the usual order $\leq$. (Hence homeomorphic when $\mathcal{V}$ is given the order topology.)
My progress: let $\mathcal{U}$ be the set of all countable unions of open intervals with rational endpoints. $\mathcal{U}$ is order-dense in $\mathcal{V}$, but it fails to be countable. I have tried and failed to reduce it to a countable set without losing order-denseness.
As for constructing a counter-example, I have not had any good ideas.
Nor have I found any useful answers on this board, nor elsewhere online.
Bonus (I conjecture the answer is 'no'):
(Q$'$) Is there a countable family $\mathcal{U}$ of open sets that is order-dense in $\tau$, the family of all open sets?
Not necessarily. Take $\mathcal V=\{V,V'\}$ where $V'=(0,1)$ and $V=(0,\frac12)\cup(\frac12,1)$. Note that if there were $U$ with $V \subsetneq U \subsetneq V'$, then $V'$ would contain at least two points that $V$ does not, but $V'$ has only one extra point. Thus (if we require strict inclusions, as in the statement of the problem) there is no family $\mathcal U$ whatsoever that is order-dense in $\mathcal V$, simply because there is no $U$ that could be inserted strictly between $V$ and $V'$ for the above choice of $\mathcal V$. (Similar reasoning for the bonus problem.)