# Given $B \subseteq \overline{A}$ how to show that every open set meeting $B$ also meets $A$.

I am trying to understand the following proof from my lecture notes for a proposition: Let $A$ be a connected subset of a topological space $X$ and suppose $A \subseteq B \subseteq \overline{A}$. Then B is connected.

The proof is as follows:

If not, then there exists open sets $U, V \subseteq X$ such that $B \cap U$ and $B \cap V$ are disjoint nonempty and $B \subseteq U \cup V$. Then $A \cap U$ and $A \cap V$ are disjoint open subsets of $A$ whose union is $A$. Since A is connected, either $A \cap U$ or $A \cap V$ must be empty; suppose that $A \cap U = \emptyset$ so that $A \subseteq V$. But since $B \subseteq \overline{A}$, every open set meeting $B$ also meets $A$; in particular $U$, (with $U \cap B \ne \emptyset$) meets $A$, a contradiction.

I cant quite graps how we know that: $B \subseteq \overline{A}$ and then subsequently how is it the case that every open set meeting $B$ also meets $A$.

Thanks.

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Let $O$ an open subset which meet $B$, and $x\in O\cap B$. Then $x\in \overline A$, and $O$ is a neighborhood of $x$, hence $O\cap A\neq \emptyset$.