There are three doors, 1, 2, and 3. You decide, arbitrarily, that you're always going to pick door 1 first - it doesn't really matter since the doors are all indistinguishable at the start.
Then there are three possible scenarios:
Scenario A:
[car] [goat] [goat]
Scenario B:
[goat] [car] [goat]
Scenario C:
[goat] [goat] [car]
These scenarios are equally probable - it's equally likely that you'll find yourself in A, B, or C, each with probability 1 in 3.
In Scenario A, you have picked door 1. Both door 2 and door 3 contain goats, so half the time door 2 is eliminated and half the time door 3 is eliminated. In both cases, you win if you stay and lose if you switch. So in Scenario A, 1 in 3 times, you lose if you switch and win if you stay.
In scenario B, you have picked door 1. Door 2 contains a car and door 3 contains a goat, so door 3 is always eliminated. If you switch you get door 2, which has a car, so in Scenario B, 1 in 3 times, you win if you switch and lose if you stay.
In scenario C, you have picked door 1. Door 3 contains a car and door 2 contains a goat, so door 2 is always eliminated. If you switch you get door 3, which has a car, so in Scenario C, 1 in 3 times, you win if you switch and lose if you stay.
As we said at the start, these scenarios are all equally likely, so switching wins 2 out of 3 times and loses 1 out of 3 times.
Technically, we've simplified the problem a bit by saying we're always picking door 1. But since the doors start out indistinguishable, this doesn't actually change the numbers at all - you could rerun the scenarios with the exact same result if you picked door 2 or door 3. In a mathematical proof, we would say "without loss of generality, we pick door 1."