The tournament is not transitive, so it contains a cycle. Any cycle of length longer than 3 has a chord; no matter which way this chord is directed, it divides the cycle into two pieces, one of which is a cycle of strictly shorter length than the original cycle. (Surely this is a classic result and has appeared before, but I was not able to find it with a cursory search.)

Yes JBL surely. However I did this without graphs.Assume that a player $x$ has lost the maximum number of times,say $j$ times.By problem there exists a player $y$ whom player $x$ has defeated.If the problem condition is to be violated then every player who has defeated $x$ has also defeated $y$.Then $y$ was defeated atleast $j+1$ times,contradicting our assumption that $x$ has lost the maximum number of times.

It's well known that every directed graph with indegree(outdegree) of every vertex at least $1$ contains a cycle.Also it is well known that if a tournoment has a directed cycle has a directed triangle so we are done.