Open Set is Union of Elements of Basis

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

Let $B$ be a basis of $T$.

Let $V$ be an open subset of $S$.

Then $V = \bigcup \left\{ {G \in B: G \subseteq V}\right\}$

Proof
Let $x$ be arbitrary.

We will prove that
 * $x \in V \implies \exists Y \in \left\{ {G \in B: G \subseteq V}\right\}: x \in Y$

Assume that
 * $x \in V$

By definition of basis:
 * $\exists F \subseteq B: V = \bigcup F$

By definition of union:
 * $\exists Y \in F: x \in Y$

By Set is Subset of Union/General Result:
 * $Y \subseteq V$

Thus by definition of subset:
 * $Y \in \left\{ {G \in B: G \subseteq V}\right\}$

Thus $x \in Y$

Assume that
 * $\exists Y \in \left\{ {G \in B: G \subseteq V}\right\}: x \in Y$

By assumption:
 * $Y \subseteq V$

Thus by definition of subset:
 * $x \in V$

Hence by definition of union:
 * $V = \bigcup \left\{ {G \in B: G \subseteq V}\right\}$