Alternative proof for closure of a connected set is connected

Question

Can someone please check this proof and if it doesn’t work why? Thank you:
Let $A$ be a connected subset and $O \subset \bar{A}$ a closed and open set. Let $B = A \cap O$. Since $O$ is an open subset of the closure it intercepts $A$ so $B \neq \emptyset$. We also have that $B$ is an open closed subset of $A$ (because $O$ is an open closed set) but $A$ is connected so $B=A \rightarrow A\subset O$. So $\bar{A} \subset O$ because $O$ is a closed set that contains $A$. By double inclusion we get that $O=\bar{A}$. We proved that $\bar{A}$ is connected.