Finding the minimum subset of intervals covering the whole set
Here's an $O(n\log n)$ algorithm, which is optimal in the algebraic decision tree model by an easy reduction from element distinctness (map each element $x$ to the interval $(x,x+\epsilon)$). We follow Kaveh's outline. Stage 1 runs in $O(n\log n)$ time, so we focus on Stage 2.

Let $I$ be the set of intervals in $C = (a,b)$. Define the intervals $l=(a,a+\epsilon)$ and $r=(b-\epsilon,b)$, and consider the interval graph over $I \cup \{l,r\}$ (two vertices are connected if the corresponding intervals intersect). I claim that every path from $l$ to $r$ corresponds to a cover of $C$, and vice versa.

First, consider a path $P$ from $l$ to $r$. Suppose that $P$ doesn't cover some point $x \in C$. We prove by induction that all the intervals in $P$ are to the left of $x$, which leads to a contradiction since $r$ is to the right of $x$. Clearly $l$ is to the left of $x$. Suppose some $p \in P$ is to the the left of $x$, and let $p'$ be the next interval on the path. If $p'$ were to the right of $x$, then since $p$ and $p'$ intersect, $x$ would have been covered. So $p'$ must also be to the left of $x$.

For the other direction, order the intervals in a cover of $C$ in increasing order of their starting point. If two consecutive intervals $(a,b),(c,d)$ do not intersect then the point $x = (b+c)/2$ is not covered. Similarly, the first interval must intersect $l$. While it is not generally true that the last interval must intersect $r$, one of the intervals must, and if we cut the cover short at the first such interval, then we obtain another cover of $C$ which also constitutes a path in the interval graph from $l$ to $r$.

We have reduced the problem of covering $C$ to finding the shortest path in an interval graph. This is solvable in time $O(n\log n)$ and space $O(n)$, as shown for example in An Optimal Algorithm for Shortest Paths on
Weighted Interval and Circular-Arc Graphs with
Applications by Atallah, Chen and Lee.