# 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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