Choose the city with a maximal number of cities accessible from it.

First solution. Number all the cities in the country as $A_1, A_2, \dots, A_n$. By the condition, there exists a city $B_2$ for which both $A_1$ and $A_2$ are accessible. Next, the cities $A_3$ and $B_2$ are accessible for some city $B_3$. Since from $B_2$ one can reach $A_1$ and $A_2$, they are also accessible for $B_3$. Continuing in the same way, we eventually find that there exists a city $B_n$ for which $A_n$ and $B_{n-1}$ are accessible. Then for $B_n$, the cities $A_1, A_2, \dots, A_{n-1}$ are also accessible, since one can reach them via $B_{n-1}$.

Second solution. Consider a city $A$ for which the maximal number of cities are accessible (if there are several such cities, pick any one of them). Suppose that some city $B$ is not accessible for $A$. Then, by the condition, there exists a city $C$ for which both $A$ and $B$ are accessible. But then for $C$, $B$ is accessible, as well as all the cities accessible for $A$. This contradicts the choice of $A$; therefore, all cities are accessible for $A$.