Solution:

We first show that there exist teams $A, B$ and $C$, such that $A$ wins over $B, B$ wins over $C$ and $C$ wins over $A$. Suppose the contrary and take the shortest cycle of $m \geq 4$ teams $A, B, C_{1}, \ldots, C_{t}$ (i.e. $t \geq 2$), such that $A$ wins over $C_{1}, C_{1}$ wins over $C_{2}, \ldots, C_{t}$ wins over $B$ and $B$ wins over $A$. Consider the game between $C_{2}$ and $A$. If $C_{2}$ is the winner then we have the desired triple and if $A$ is the winner then we have a shorter cycle.

Now we shall use induction on $k$. The case $k=3$ was considered above. Suppose that the teams $A_{1}, A_{2}, \ldots, A_{k}$ satisfy the condition of the problem for $3 \leq k < n$. There are two cases to be considered.

Case 1. There exists a team $U \notin \{A_{1}, A_{2}, \ldots, A_{k}\}$, for which there are two teams $A_{i}$ and $A_{j}$ such that $A_{i}$ wins over $U$ and $U$ wins over $A_{j}$. Without loss of generality assume that $A_{1}$ wins over $U$. Let $A_{\ell}$ be the team of the least index that loses from $U$. Then the following $k+1$ teams
$$A_{1}, \ldots, A_{\ell-1}, U, A_{\ell}, \ldots, A_{k}$$
have the desired property.

Case 2. For any two teams $A_{i}$ and $A_{j}$ and any team $U$ either $A_{i}$ loses from $U$ or $A_{i}$ and $A_{j}$ both win over $U$.

Partition all teams apart from $A_{1}, A_{2}, \ldots, A_{k}$ in two sets $S$ and $T$, such that all teams from $S$ win over $A_{1}, A_{2}, \ldots, A_{k}$, and all teams from $T$ lose from $A_{1}, A_{2}, \ldots, A_{k}$. It is clear that $S \cap T = \varnothing$ and none of $S$ and $T$ is the empty set.

Let $U \in S$ and $V \in T$ be such that $V$ wins over $U$ (such a pair exists due to the condition of the problem). Now the following $k+1$ teams
$$U, A_{1}, \ldots, A_{k-1}, V$$
have the desired property.