Showing that tournament sort requrires O(n log n) comparisons
Initially you place the elements you want to sort in the leaves of the tournament tree. Then you fill all the internal nodes with the bigger of the two elements in their respective children. This takes $n-1$ comparisons.

After that, you have the largest element in the root. So you can remove it and place it in the output. Now all the comparisons where this element was involved have to be redone. This is one comparison per level of the tree, making $(\log n) -1$ comparisons (-1, since on the lowest level, there is only one candidate element left, so you don't need to compare.)

Now the second largest element is at the top and you can repeat the procedure. And so on. In the end you will have made up to $\log n$, i.e. $O(\log n)$ comparisons for each of the $n$ elements.

Summing up, we have $n-1 + n\cdot O(\log n)$ comparisons. By the rules of $O$-notation, that is in $O(n\log n)$.