Proof by induction
Suppose a tournament of $n$ players have a dominant player (this is trivially true for $n=1,2,3$). We now try to prove that a tournament of $n+1$ players also have a dominant player, which is either the previous dominant player or the new player himself.
Let's call the previous dominant player $D$ and the new player $N$.
Suppose $D$ beats $N$. Then $D$ is still a dominant player in the tournament of $n+1$ players.
Then now we can assume that $N$ beats $D$.
Now, if there is a person $H_1$ such that $D$ beats $H_1$ and $H_1$ beats $N$, then $D$ is still a dominant player, since it dominates all other player, including the new player $N$.
So this means now we can assume that $N$ beats everyone that was beaten by $D$.
This means everyone that was dominated by $D$ will be dominated by $N$ also. And since $N$ beats $D$, $N$ dominates the whole tournament, making $N$ the new dominant player, proving the claim. =)
More formally, we'll represent the players as vertices in a graph, and the matches outcome as directed edges between those vertices.
First, some notations:
Let $G = \{V, E\}, V = \{v_1, v_2, \ldots, v_n\}, E = \{(v_i, v_j) | \text{player }i\text{ beats player }j\}$
Let $d_G(v_i, v_j)$ be the shortest path length from $v_i$ to $v_j$ in $G$.
Let $s_G(v)$ be the maximum distance from $v$ in $G$ (formally: $s_G(v) = max(\{d_G(v, v_i)\ |\ \forall v_i \in G\})$)
To be proven:
In any graph $G$, $min(\{s_G(v)\ |\ v\in G\}) \leq 2$
We'll prove by induction.
Let $n = |V|$, which is the number of vertices in $G$.
For $n=1,2,3$, it's trivial, since the longest path in such graphs will be 2.
Assume the property holds for $|V| = n$. Now we are trying to prove that it also holds for $|V| = n+1$.
Let $G$ be such that $|V| = n$. By induction hypothesis, there is a vertex with distance at most 2 to any other vertices. Without loss of generalization, $s_G(v_1) \leq 2$
We want to add one new vertex $v_{n+1}$ to $G$ and prove that the property still holds.
Now, if $s_G(v_1) = 1$ (i.e., player $1$ beats all other players in $G$), we have two cases:
$(v_{n+1}, v_1) \in E$ (i.e., player $n+1$ beats player $1$)
This implies $s_G(v_{n+1}) = 2$, so the property still holds.
$(v_1, v_{n+1}) \in E$ (i.e., player $1$ beats player $n+1$)
This implies $s_G(v_1) = 1$, so the property still holds.
So we can now assume that $s_G(v_1) = 2$
If $(v_1, v_{n+1}) \in E$, then we still have $s_G(v_1) = 2$ and the property still holds.
So we can now assume that $(v_{n+1}, v_1) \in E$
Let $H_1, H_2$ denotes, respectively, the vertices of distance 1 and distance 2 from $v_1$.
$$
H_1 = \{v_i\ |\ d(v_1, v_i) = 1\}\\
H_2 = \{v_i\ |\ d(v_1, v_i) = 2\}
$$
Note that $H_1 \cup H_2 = V-\{v_1\}$ since $s_G(v_1) = 2$
If $v_{n+1} \in H_2$, we are done, since we'll still have $s_G(v_1) = 2$.
So we can now assume that $v_{n+1} \not\in H_2$
Note that for any vertices $v_i \in H_2$, we have $(v_i, v_1) \in E$, because otherwise $d(v_1, v_i) = 1$. Also note that $$v_i \in H_2 \Rightarrow \exists v_j \in H_1, (v_j, v_i)\in E \qquad \ldots\ (1)$$ (i.e., if player $i$ is of distance 2 from player $1$, there exists another player that is beaten by player $1$ and that player beats player $i$).
Since $v_{n+1} \not\in H_2$, it means $$\forall v_j\in H_1, (v_{n+1}, v_j)\in E \qquad \ldots (2)$$ (i.e., everyone that was beaten by player $1$ is also beaten by player $n+1$).
By $(1)$ and $(2)$, we have $$\forall v_i \in H_2, d(v_{n+1}, v_i) \leq 2$$ (i.e., everyone with distance 2 from player $1$ is at most of distance 2 from player $n+1$)
This means we have $$\forall v_i \in H_1, d(v_{n+1}, v_i) = 1\\\forall v_i \in H_2, d(v_{n+1}, v_i) \leq 2$$ which means that $s_G(v_{n+1}) \leq 2$ (since $H_1 \cup H_2 = V-\{v_1\}$), so the property holds.
We have proven that if the property holds in a graph with $n$ vertices, it also holds in a graph with $n+1$ vertices. By induction, the property is true for all $n$.
Q.E.D.