Python

Complexity O(nlog n)

#!/bin/env python3

n = int(input())
trips = []
for _ in range(n):
    times = input().split(' ')
    trips.append((int(times[0]), int(times[1])))

# Sort by decreasing arrive_time when the leave_time is the same
trips.sort(key=lambda x:x[1], reverse=True)
# By increasing leave_time
trips.sort(key=lambda x:x[0])

# Remove duplicates
trips = [trips[0]] + [b for a,b in zip(trips, trips[1:]) if a!=b]
stack = []
for trip in trips:
    while stack and stack[-1][1] >= trip[1]:
        #The trip on the stack is worse than trip
        stack.pop()
    stack.append(trip)

print(len(stack))
for a, b in stack:
    print(a, b)

Explanation

The stack contains trips not yet determined to be bad. The trips are processed in ascending leave_time. This means any trip already on the stack, has a lower or equal leave_time than the trip we are currently looking at.

If the trip on the top of the stack arrives later than the current trip,
that means the trip on the stack starts earlier and ends later
than the current trip. Therefore the stack trip is worse, and is popped off the stack.

This is repeated until the trip on the top of the stack ends before the current trip.
At that point, all trips down the stack are safe, because
no trip further down the stack ends after the top trip.
After the stack-popping is done, we add the current trip.

If multiple trips start at the same time, the one
with the larger arrive_time comes first, because
it needs to be popped by the better trip, when it comes later.
The sorting we do at the start makes sure
that a trip always comes before the trip it is objectively worse than.
This means the last trip is guaranteed to be good.