Intuition of lower bound for finding the minimum of $n$ (distinct) elements is $n-1$ as dealt with in CLRS
Every match has exactly one winner and one loser. If every element except the winner must lose at least one match, there must be at least $n-1$ matches, as otherwise there would be a match with two losers. So, at least $n-1$ matches are nessecary to determine a winner, i.e. $n-1$ comparisons are nessecary to determine the minimum.

The explanation given in the book is valid, but I personally prefer another approach. (which does require some elementary graph theory) Suppose you draw the graph with the elements as nodes and you draw an edge for each pair of elements that your algorithm compares. Note that it is impossible to know which of a pair of elements is smaller if there is no path between them in this graph. So, if we have determined the minimum, this graph must connected. Any connected graph on $n$ nodes has at least $n-1$ edges, so there must have been $n-1$ comparisons.