 Solution

Let $I_i=[s_i,t_i]$ be an interval whose earliest end triggers greedy meeting $i$. After that meeting, every remaining interval starts strictly after $t_i$: its end is at least $t_i$, and otherwise it would contain $t_i$ and be removed. HenceThe witness intervals $I_1,\dots ,I_k$ are therefore pairwise disjoint. Every feasible schedule needs a distinct meeting in each one, yielding a lower bound of $k$. The greedy schedule uses exactly $k$ meetings and covers everyone, so it is optimal. This is an optimality certificate for the interval stabbing problem.

To follow the source's suggested contradiction explicitly, suppose a schedule $u_1<\cdots <u_j$ with $j<k$ covered everyone. Since $I_1$ ends at $t_1$, some $u_{\ell }\le t_1$, so $u_1\le t_1$. If $u_i\le t_i$ and $i<j$, then none of the first $i$ alternative meetings can hit $I_{i+1}$, which starts after $t_i$. A later meeting must hit it by its end $t_{i+1}$, implying $u_{i+1}\le t_{i+1}$. Mathematical induction gives $u_i\le t_i$ for $1\le i\le j$. But $I_{j+1}$ starts after $t_j\ge u_j$, later than every alternative meeting. It is uncovered, a contradiction. For $j=0$ a nonempty family is immediately uncovered. The source's hint starts at index zero without defining $u_0,t_0$; the proof above starts at the actual first meetings instead.