For this question, the most straight-forward way is probably to think
about the decision tree for 2nd-smallest and the lower bound for that.

With n=5, we know that to find the 2nd-smallest we need to:
   - find the smallest, which costs 4 comparisons
   - the decision tree we draw if we do the minimum finding
       algorithm where we compare pairs and only the smallest
       of each pair makes it to the next "round" shows that
       there are only 3 candidates left for 2nd-smallest, and
       finding the smallest of those 3 costs 2 comparisons

Now you've used up 6 comparisons.  To do better than 8 you can only use
one more comparison.

So, if you can use the decision tree logic to show there are at least 3 
candidates for 3rd-smallest at this point, you've got an overall lower-bound
of 8 proven.