Relation between the "Point-Cover-Interval" problem and the "Interval Scheduling" problem
Here is a proof that the two numbers are equal:


  For every interval $I$ consider the largest set of disjoint intervals
  with $I$ as the rightmost interval. This has a size, say $D_I$.
  
  Now if $D_I = D_J$, then $I$ and $J$ have to intersect, so the set of
  intervals with equal $D_I$ values all pairwise intersect each other, and so have a
  common point (by the 1D version of Helly's theorem, or by just considering the intervals arranged in order of their left endpoints).
  
  Now if $d = \max D_I$ (the size of the largest set of non-intersecting
  intervals), then we have $D_I$ taking all the values $\{1, 2, \dots, d\}$,
  and thus $d$ points are enough to cover all the intervals, and we need at least $d$ to cover the largest set of disjoint intervals.


For some related reading, look at Mirsky's theorem, which is a dual of Dilworth's theorem.


  Dilworth's theorem says that in a set of elements with a partial order defined on them, the size of the largest antichain
  (antichain is a set of elements not comparable to each other) is same
  as the size of the smallest partition into chains (each set of the
  partition is a total ordered set).
  
  Mirsky's theorem says that the size of the largest chain is the same as the smallest number of antichains into which the set might be
  partitioned.


In this case defines your order relation as follows:

$$I = (i,j) \lt I' = (i', j') \iff j \lt i'$$

Basically, two disjoint intervals are comparable, with the left one being smaller.

Because of Helly's theorem, the two numbers you are computing are exactly the numbers in Mirsky's theorem, which is defined as the height of the the partial order (Dilworth's theorem defines the width).