Make a graph where two contestants are connected with an edge if they $\textbf{did not}$ play each other. Now we have that whenever there is an edge $AB$ in the graph, there is also an edge $CD$ such that $AC$, $AD$, $BC$ and $BD$ aren't connected with edges. Call such pairs of edges $\textit{conjugates}$. Also any vertex that is not an endpoint of the conjugate edge $CD$ is connected to at least one of $A$, $B$. Now we see that if the graph isn't just one pair of conjugate edges then it must be connected. Take one of the vertices with maximal degree $X$. Now take a vertex $Z$ which is an endpoint of the conjugate of an edge $XU$. Notice for each neighbour $Y$ of $X$, either $ZY$ is an edge, or $Z$ is the endpoint of the conjugate edge of $XY$. This means $Z$ has at least as many neighbours as $X$, so it too has maximal degree. Now same reasoning gives $U$ has maximal degree, and now we can prove every vertex has maximal degree using breadth first search, since the graph is connected.