Finding the minimum subset within a cycle
Using Kaveh's reduction in this related question, we only have to solve the case where the given intervals cover the entire circle $[c,d]$.

Here's an algorithm that almost calculates the optimum, in $O(n\log n)$. Replace each interval $[a,d] \cup [c,b]$ by the two intervals $[a,d],[c,b]$, and now solve the original problem. Let $O$ be the optimum of the original problem, and $S$ be the optimum of the new problem. I claim that $S-1 \leq O \leq S$.

First, given any solution to the new problem, we can find a solution to the old problem with the same cost, showing that $O \leq S$. For the other direction, take some optimal solution. If it doesn't use any split interval, then $S = O$. Otherwise, let $I_\min$ be the split interval minimizing $a$, and let $I_\max$ be the split interval maximizing $b$. If we drop all other split intervals, then whatever remains is still a cover of the circle. If $I_\min \neq I_\max$ then $S = O$, and otherwise all we know is that $S \leq O+1$.

It's plausible that this algorithm can be adapted somehow to actually find the optimum. (Most algorithms that work on interval graphs also work on circular-arc graphs.)