Consider K=1 (2 players), the probability is 1.
Consider K=2 (4 players); first round can be in following ways:
1-2 | 3-4
1-3 | 2-4
1-4 | 2-3
They don't meet in finals in first case (of 3). Thus, for K=2, the probability that they meet is 2/3.
I assure you that this probability gets close to 1/2 for large K. Just notice that if 1 and 2 meet before final, they dont meet at final.

N/2(N-1)

If you worked out conditional probability, that is fine. Here we present the combinatorial solution. Imagine the levels proceed in any random way such that player X and Y appear for final level. This can be imagined as N players getting partitioned into two groups of N/2 players each, with player X topping in first group and player Y in second. The best player of each partition reaches final. We only need to ensure that above partition separates player 1 from 2. The probability that a random partition separates 1 & 2 is (N/2) / (N-1) (how?)

Here is how: for creating a partition to separate them, we pick (N/2) people from N, wishing to take player 1, but not player 2.