A problem of non-emptiness of intersections of certain chains of regular open sets Let $X$ be a topological space and $\mathrm{RO}(X)$ its complete boolean algebra of regular opens. Define well inside relation: $$U\prec V\iff\overline{U}\subseteq V.$$ Let $\mathcal C\subseteq\mathrm{RO}(X)$ be a chain such that (a) any two distinct elements of $\mathcal C$ are comparable w.r.t. $\prec$ and (b) $\mathcal C$ is strictly descending, i.e. for every $U$ in the chain there is $V$ in the same chain with $V\subsetneq U$.
Of course, the intersection of $\mathcal C$ does not have to be non-empty (e.g. take $\{(n,+\infty)\mid n\in\omega\}\subseteq\mathrm{RO}(\mathbf{R})$, with $\mathbf{R}$ the set of reals), but it is always non-empty if the space is compact.
Say that a chain $\mathcal  C$ covers $\mathcal D$ iff for every $U\in\mathcal C$ there is $D\in\mathcal D$ such that $D\subseteq U$. Further, call $\mathcal C$ c-minimal if for every $\mathcal D$ covered by $\mathcal C$, $\mathcal C$ is also covered by $\mathcal D$.
My problem is whether for every non-empty c-minimal chain $\mathcal C$ (in an atomless $\mathrm{RO}(X)$) which satisfies (a) and (b), $\bigcap\mathcal C$ is non-empty.
EDIT: Changed the terminology since, as pointed in the comments, the one I used originally was a bit misleading. 
 A: Here is a provisional negative answer.
If $\mathcal{C}$ is a c-minimal chain then $N=\bigcap\mathcal{C}$ is closed and nowhere dense (if nonempty): if the interior is nonempty take a nonempty regular open set $O$ such that $\overline{O}\subseteq \operatorname{int}N$. Then the chain $\mathcal{C}\cup\{O\}$ is covered by $\mathcal{C}$ but it does not cover it. This shows that a c-minimal chain does not have a minimum and that $\bigcap\mathcal{C}=\bigcap\{\overline{C}:C\in\mathcal{C}\}$ is closed.
Now the chain $\{C\setminus N:C\in\mathcal{C}\}$ is c-minimal in $X\setminus N$ and its intersection is empty.
This is provisional in the sense that I could not think of a c-minimal chain. For example, in the real line every chain is countable and by diagonalising a co-initial sequence one can construct a strictly smaller chain. Correction: every well-ordered (up or down) chain is countable; every chain still has a co-initial sequence.
Added 2023-01-09: Consider the ordinal $\omega_1$ with its order topology. It is well known that the intersection of two closed and unbounded sets is again closed and unbounded. From this we deduce that if $U$ and $V$ are open such that $U\prec V$ then either $\overline{U}$ is compact or $V$ contains an interval $[\alpha,\omega_1)$ for some $\alpha<\omega_1$. For is $\overline{U}$ is not compact then it is unbounded; the complement of $V$ is closed and disjoint from $\overline{U}$, and hence bounded.
The chain $\mathcal{C}=\bigl\{[\alpha+1,\omega_1):\alpha<\omega_1\}$ consists of clopen sets and it satisfies conditions (a) and (b). If $\mathcal{C}$ covers a chain $\mathcal{D}$ then all members of $\mathcal{D}$ are unbounded and the comment above shows that every member of $\mathcal{D}$ contains a member of $\mathcal{C}$.
Thus $\mathcal{C}$ is c-minimal and its intersection is empty.
