Well, the second line replaces the first line 50% of the time.
3 lines: The 3rd line has the correct 1/3 chance of being written, otherwise (with probability 2/3) one of the first 2 lines remain, and they had an equal 50% chance of being stored just prior to the 3rd line appearing, which now is 50% x 2/3 = 1/3.
4 lines: The 4th line has the correct 1/4 chance of being written, otherwise (with probability 3/4) one of the first 3 lines remain, and they had an equal 1/3 chance of being stored just prior to the 4th line appearing, now 1/3 x 3/4 = 1/4. Same argument applies all the way down. That's very neat! ...
n lines: The nth line has the correct 1/n chance of being written, otherwise (with probability (n-1)/n) one of the first n-1 lines remain, and they had an equal 1/(n-1) chance of being stored just prior to the nth line appearing, now 1/(n-1) x (n-1)/n = 1/n.
3 lines: The 3rd line has the correct 1/3 chance of being written, otherwise (with probability 2/3) one of the first 2 lines remain, and they had an equal 50% chance of being stored just prior to the 3rd line appearing, which now is 50% x 2/3 = 1/3.
4 lines: The 4th line has the correct 1/4 chance of being written, otherwise (with probability 3/4) one of the first 3 lines remain, and they had an equal 1/3 chance of being stored just prior to the 4th line appearing, now 1/3 x 3/4 = 1/4. Same argument applies all the way down. That's very neat! ...
n lines: The nth line has the correct 1/n chance of being written, otherwise (with probability (n-1)/n) one of the first n-1 lines remain, and they had an equal 1/(n-1) chance of being stored just prior to the nth line appearing, now 1/(n-1) x (n-1)/n = 1/n.