Union of Local Bases is Basis

Theorem
Let $T = \left({X, \tau}\right)$ be a topological space.

Let $\mathcal B_x$ be a local basis at a point $x \in X$ which consists entirely of open sets.

Then $\displaystyle \mathcal B = \bigcup_{x \in X} \mathcal B_x$ is a basis for the topology $\tau$.

Proof
Let $U \in \tau$ be any open set of $T$.

Consider any $x \in U$.

Then, by definition of local basis, $\exists B_x \in \mathcal B_x$ such that $B_x \subset U$.

Also by definition of local basis $x \in B_x$.

So then:
 * $\displaystyle \bigcup_{x \in U} \left\{{x}\right\} = U \subseteq \bigcup_{x \in U} B_x$

Also, since $B_x \subset U$:
 * $\displaystyle \bigcup_{x \in U} B_x \subseteq U$

Thus:
 * $\displaystyle U = \bigcup_{x \in U} B_x$

So, by definition, $\displaystyle \mathcal B = \bigcup_{x \in X} \mathcal B_x$ is a basis for the topology $\tau$.