Let $X$ be a topological space and $\mathcal{B}$ a basis for the topology on $X$. Then $U$ is open in $X$ if and only if for each $p \in U$ there exists $B \in \mathcal{B}$ such that $p \in B \subseteq U$.
Proof. Assume $U$ is open. Since $\mathcal{B}$ is a basis, we have $$U = \bigcup_{\alpha \in A} B_\alpha$$ where $B_\alpha \in \mathcal{B}$ for each $\alpha \in A$. Thus for each $p \in U$ we have $p \in \bigcup_{\alpha \in A} B_\alpha$ and so $p \in B_\alpha$ for some $\alpha \in A$. But since also $\bigcup_{\alpha \in A} B_\alpha \subseteq U$ we have $B_\alpha \subseteq U$. Conversly, we can write $$U = \bigcup_{p \in U} B_p$$ for some $U \subseteq X$ where for each $p \in U$ we have $B_p \in \mathcal{B}$. Since each basis element is open, $U$ is open as a union of open sets. $\Box$
Is this correct?