A problem about open cover of connected space Let $\mathscr{B}$ be an open cover of connected space $X$，and $\varnothing\notin \mathscr{B}$.
For any $U,V\in \mathscr{B}$, does there exist a finite sequence $U,A_2,A_3,\cdots,A_{n-1},V$ in $\mathscr{B}$ such that the adjacent two open sets are intersected?
 A: Yes.
Call $U$ and $V$ "finitely connected" if the desired property is true, and denote $F_U = \{V \in \mathscr{B} \mid $U$ \text{ and } $V$ \text{ are finitely connected}\}$. We want to show $F_U = \mathscr{B}$. Define:
$$A_U = \bigcup_{V \in F_{U}}V; \qquad \qquad B_U = \bigcup_{V \in \mathscr{B} \backslash F_U}V$$
Then $A_U$ and $B_U$ are clearly open, and $A_U \cup B_U = X$ since $\mathscr{B}$ is an open cover. We want to show that $A_U \cap B_U = \emptyset$. Suppose not: Then there would exist $V \in F_U$ and $W \in \mathscr{B} \backslash F_U$ such that $V \cap W \neq \emptyset$. But since $V$ is finitely connected to $U$, we have a sequence $U, A_1, \ldots, A_{n-1}, V$ as described in the problem statement. Then we can add $W$ onto the end of this sequence, and this means $U$ and $W$ are finitely connected as well, so $W \in F_U$, which is a contradiction.
Finally, since $X$ is connected, we have $A_U = \emptyset$ or $B_U = \emptyset$. Clearly $U \subset A_U$, and $U$ is not empty since $\emptyset \notin \mathscr{B}$. So $B_U = \emptyset$, which means there are no sets in $\mathscr{B} \backslash F_U$, ie $F_U = \mathscr{B}$.
