Proof
(related to Proposition: A Necessary Condition of a Neighborhood to be Open)
 By hypothesis, $B$ is a neighborhood of all of its points in a topological space $(X,\mathcal O).$
 Then, by definition of neighborhood, for every $A_i\in B$ there is an open set $O_i\in\mathcal O$ with $A_i\in O_i\subseteq B$ ($i\in I$, $I$ being an appropriate index set).
 Therefore, $B$ can be expressed as the set union of all these open sets. $B=\bigcup_{i\in I} O_i.$
 By the third axiom of topology, $B$ is open.
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References
Bibliography
 Steen, L.A.;Seebach J.A.Jr.: "Counterexamples in Topology", Dover Publications, Inc, 1970
 Jänich, Klaus: "Topologie", Springer, 2001, 7th Edition