It's pretty intuitive if you increase it to 100 doors.
You choose 1 door out of 100. The chance of getting it right is 1/100 or 1%.
The host offers to close 98 wrong doors and leave one door left. This set of 99 doors has 99/100 or 99% chance of being right.
Monty opens the 98 of 99 doors but they still belong to the set of 99% probability, so the one door lefts inherits that probability as 98 goats have been removed.
Now just reduce it to 3 doors.
You pick one, probability is 1/3 or 33.333% of it being correct.
Monty now offers the other set of doors containing 2/3 or 66.666% of being correct.
He opens one of the doors, but he ALWAYS opens the goat one. If he did it randomly it would be 50/50. But because he chooses a set where he always opens the goat ones it keeps the set probability of 2/3'rds.