Solution:
Every point must have at least one edge. We show that there is a point with just one edge. Suppose the contrary, that every point has at least two edges. We now construct a path in which the same edge or point never appears twice. Starting from any point $b$, move along an edge to $c$. $c$ is not already on the path, because otherwise the edge would join $b$ to itself. Now suppose we have reached a point $x$ not previously on the path. $x$ has at least two edges, so it must have another one besides the one we used to reach it. Suppose this joins $x$ to $y$. If $y$ is already on the path, then we have two distinct ways of moving along edges from $x$ to $y$: directly, or by backtracking along the path from $x$ to $y$. But this is impossible, so $y$ is not already on the path and we may extend the path to it. But this procedure allows us to construct a path containing more than the $n$ distinct points available. Contradiction.

The result is now easy. Induction on $n$. Take a point with just one edge. Remove it and the edge. Then the remaining $n - 1$ points satisfy the premise and hence have just $n - 2$ edges.