The answer by Aritro works and is more elegant, but if you must use induction, there's a way to do it using strong induction.
Label the city $A_1, \dots A_n$, and suppose $A_1$ is the city that can be reached from any other city through at most 1 other city. And now we add $A_{n+1}$. If the road between $A_1$ and $A_{n+1}$ leads to $A_1$ then we are done. So now we assume that road leads to $A_{n+1}$.
Divide the cities $A_2, \dots, A_n$ into three categories:
Category 1: the ones that lead directly to $A_{n+1}$.
Category 2: the ones where $A_{n+1}$ leads into it, but it leads into $A_1$
Category 3: the ones where $A_{n+1}$ leads into it, and $A_1$ also leads into it.
Obviously from category 1 we can reach $A_{n+1}$ easily. From category 2, we can also reach from that city through $A_1$ then to $A_{n+1}$.
If category 3 is empty, then $A_{n+1}$ is our required city, because from all the remaining cities ($A_1$, category 1, and category 2) we can reach $A_{n+1}$ through at most one other city.
So now we assume that category 3 is not empty.
Case 1: category 3 contains multiple cities.
We invoke the induction hypothesis one more time, ONLY on category 3 cities. There is a city say $B$ that satisfies the required property among category 3 cities. Then we assert that $B$ is our required city for all $n+1$ cities.
Meaning, from other cities in category 3, we can reach $B$ through at most 1 other city by the induction hypothesis. From cities in category 1, we can reach $B$ through $A_{n+1}$. From cities in category 2, we can reach $B$ through $A_1$.
Case 2: category 3 contains a single city, say $B$.
We cannot invoke the induction hypothesis because the premise requires there are at least 2 cities, but we can still assert that $B$ is our required city. From any other cities ($A_1, A_{n+1}$, category 1, and category 2) we can reach $B$ either directly or through $A_1$ or $A_{n+1}$.