Greedy algorithm for maintaining drugs
Here is a counterexample.
There are 6 drugs with intervals $$[1, 5],\ [2,8],\ [3,8]$$
and $$[6, 11], \ [6,12],\ [9, 13].$$
We can use containers of volume 4 to cover the first 3 drugs and containers of volume 10 to cover the last 3 drugs. So it is enough to use 2 kinds of containers.
If we use the greedy algorithm, we would first cover $$[2,8],\ [3,8],\ [6, 11], \ [6,12]$$
using, for example, containers of volume 7. And then the remaining two intervals $$[1,5],\ [9,13]$$
that are disjoint will need additional two containers of different volumes to cover.