We say that two groups of islands $\mathcal{A}$ and $\mathcal{B}$ form a network if and only if for any $A \in \mathcal{A}$ and $B \in \mathcal{B}$, there is a route between $A$ and $B$.

1. First choose any route, and let its endpoints be islands $A$ and $B$. Since $n \ge 3$, there must exist a third island $C$; and by the assumption of the problem, it must be connected to $A$ or $B$, so without loss of generality assume it is connected to $A$. Define $\mathcal{A}_0 = \{A\}$, $\mathcal{B}_0 = \{B, C\}$; it is easy to see that $(\mathcal{A}_0, \mathcal{B}_0)$ forms a network. Note that $|\mathcal{A}_0 \cup \mathcal{B}_0| = 3$.

2. Next we prove that: if $|\mathcal{A}_0 \cup \mathcal{B}_0| \ge 3$, then in every year, we can always redivide $\mathcal{A}_0 \cup \mathcal{B}_0$ into two nonempty groups of islands $\mathcal{A}_1$ and $\mathcal{B}_1$, such that $(\mathcal{A}_1, \mathcal{B}_1)$ forms a network.
If the route closed by the company does not link $\mathcal{A}_0$ and $\mathcal{B}_0$, then we may simply keep the original grouping. Now suppose the company has closed some route linking $A \in \mathcal{A}$ and $B \in \mathcal{B}$. Let us consider $\mathcal{A}_1 = \{A, B\}$ and $\mathcal{B}_1 = (\mathcal{A}_0 \cup \mathcal{B}_0) - \{A, B\}$; it is easy to see that $\mathcal{A}_0 \cup \mathcal{B}_0 = \mathcal{A}_1 \cup \mathcal{B}_1$, and both $\mathcal{A}_1$ and $\mathcal{B}_1$ are nonempty (since $|\mathcal{A}_0 \cup \mathcal{B}_0| \ge 3$).

Below we show that $(\mathcal{A}_1, \mathcal{B}_1)$ forms a network. Consider $C \in \mathcal{A}_0 - \{A\}$; since $(\mathcal{A}_0, \mathcal{B}_0)$ forms a network, there must be a route between $C$ and $B$. If there was originally no route between $C$ and $A$,

then when $\overline{AB}$ is closed at the end of the year, $\overline{AC}$ must be established. In other words, $C$ must have a connection to both $A$ and $B$. Similarly, for all $C \in B_0 - \{B\}$, $C$ must have a connection to both $A$ and $B$. This means that any island in $B_1$ has a connection to both $A$ and $B$, so $(A_1, B_1)$ forms a network. This completes the proof.

3. Suppose $|A_1 \cup B_1| < n$. We now prove that: after some number of years, we can find an island $C$ not in $A_1 \cup B_1$, and redivide $A_1 \cup B_1 \cup \{C\}$ into two nonempty groups of islands $A_2$ and $B_2$, such that $(A_2, B_2)$ forms a network.
Let $D \notin A_1 \cup B_1$. By the assumption of the problem, after some number of years, the company must close the route between $D$ and some island in $A_1 \cup B_1$, without loss of generality say it is $A \in A_1$. Note that $(A_1, B_1)$ forms a network, so all points in $B_1$ have a route to $A$; therefore, when $\overline{AD}$ is closed, all points in $B_1$ will have a route to $D$. Therefore, consider $A_2 = A_1 \cup \{D\}$, $B_2 = B_1$, then $(A_2, B_2)$ forms a network and $A_2 \cup B_2 = A_1 \cup B_1 \cup \{D\}$. This completes the proof.

4. By 3., we may assume that after some number of years, we can divide all $n$ islands into two nonempty groups of islands $A$ and $B$, such that $(A, B)$ forms a network. Now suppose that after some number of years, the route between $A \in A$ and $B \in B$ is closed. Let $A' = \{A, B\}$, and let $B'$ be all the other islands, then it is easy to see that $(A', B')$ forms a network.

Suppose that after some further number of years, without loss of generality, the route between $A$ and some $B' \in B'$ is closed. Since $B$ has a route to all islands in $B'$, this means that when $\overline{AB'}$ is closed, $\overline{AB}$ will be established, that is, $B$ has a route to every other island. This completes the proof.