Interval schedule maximization when encountering intervals with equivalent end points
The greedy algorithm can be characterized as follows:

Let $I$ be the initial set of intervals 

1) Select an interval, $x \in I$, with the earliest finishing time.

2) Remove $x$ and all intervals intersecting $x$ from $I$, obtaining $I^{'}$.

3) add $x$ to the solution

4) Set $I=I^{'}$

5) Continue until $I$ is empty.

Here, $x$ is some remaining interval that finishes before all other intervals in $I$. When we add $x$ to our solution we have already removed any interval with the same finish time as $x$ that may have had a start time which was "too early." If there is some other $y\in I$ with the same finish time as $x$ that is present when $x$ is chosen, then by necessity, both $x$ and $y$ begin after the last interval we added to our solution, so either is a legitimate choice.