Q:

There is a best performer in a round robin tournament

At a social bridge party every couple plays every other couple exactly once. Assume there are no ties. If $n$ couples participate, prove that there's best couple in the following sense: A couple $u$ is best if for every couple $v$, $u$ beats $v$ or $u$ beats a couple that beats $v$.

What I tried:
If $u$ beats some $v$, the base holds.
Suppose $u$ beats $k$ different $v$'s. Then if $u$ beats $(k + 1)$st $v$, we are done. Suppose $u$ loses to $(k + 1)$st $v$, then if one of the $k$ $v$'s beats $k + 1st$ $v$, we are done. Suppose that doesn't happen. Then $(k + 1)$st $v$ will have beaten $k$ $v$'s by transitivity since $(k + 1)$st $v$ beats $u$, $k + 1$ in total. This shows the existence of best couple.
Can you check my work, please? 

A:

This argument does not work.  You may not assume that $u$ beats $k$ different $v$'s.  All you may assume is that for $k$ different $v$'s, either $u$ beats directly or "transitively".  Unfortunately when the $k+1$st $v$ comes along, the transitivity is now two steps long and not one step, so the induction breaks down.
Here's a solution, which has some of the same flavor as the original attempt. Write $xBy$ to denote $x$ beats $y$ directly.  Since each couple plays every other couple, exactly one of $xBy$ or $yBx$ must hold.  For every couple $x$, define $X'=\{y:xBy\}$ and $X''=\{z:\exists y\in X', yBz\}$.  Define $f(x)=|X'\cup X''|$.  Note that $\{x\}, X'\cup X''$ are disjoint, and we are trying to prove that there is some couple $x$ such that $f(x)=n-1$.
Okay, let $u$ be chosen so that $f(u)$ is maximal.  If $f(u)=n-1$, we are done; otherwise choose any $v\notin \{u\}\cup U'\cup U''$.  Our strategy is to prove $f(v)>f(u)$.  
Lemma 1: $U'\subseteq V'$.  Proof: If $t\in U'\setminus V'$, then $uBt$ and $tBv$, which contradicts $v\notin U''$.
Lemma 2: $U''\subseteq V''$.  Proof: If $t\in U''$, then there is some $s\in U'$ and $sBt$.  But by Lemma 1, $s\in V'$, so in fact $t\in V''$.
Note also that $u\in V'$, but $u\notin U'\cup U''$.  Hence $$f(v)=|V'\cup V''|\ge |\{u\}|+|U'\cup U''|=1+f(u)$$
But now $f(v)>f(u)$, which contradicts the maximality of $f(u)$.

