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.