Statement
Let be $X$ a topological space that has a open-closed base $\mathcal{B}$. So if $X$ is such that $|X|>1$ then it is not locally connected.
Proof. If $X$ was locally connected then for any $x\in X$ there exist a connected neighborhood $V$ of $x$ and so there must exist $B\in\mathcal{B}$ such that $x\in B\subseteq V$ and so, since $B$ is open and closed, $B$ and $V\setminus B$ is two open disjoint set of $V$ such that $B\cup V\setminus B=V$, that means that $V$ is not connected.
Clearly the proof is correct iff $B\subset V$: indeed only in this case it follow that $V\setminus B\neq\varnothing$. Anyway I see that any discrete space $X$ is a space such that have a open-closed base and it is locally connected: so it seems that the statement is false; however my text says the contrary so I ask if we add other hypothesis then the statement is true. So how to prove the statement? Could someone help me, please?
For each $x\in X$ let $\mathscr{B}(x)=\{B\in\mathscr{B}:x\in B\}$; it’s enough to add the requirement that there be a point $x_0\in X$ such that $\bigcap\mathscr{B}(x_0)$ is not open. (This is automatic if, for instance, $X$ is $T_1$ but not discrete.)
Let $H=\bigcap\mathscr{B}(x_0)$, and let $V$ be an open nbhd of $x_0$; then $H\subsetneqq V$, so fix $x\in V\setminus H$. There are $B_0,B_1\in\mathscr{B}(x_0)$ such that $B_0\subseteq V$ and $x\notin B_1$. Let $U=B_0\cap B_1$; then $U$ is clopen, and $x_0\in U\subseteq V\setminus\{x\}$, so $U$ and $V\setminus U$ are disjoint, non-empty open subsets of $V$ whose union is $V$.