Existence of Subfamily of Cardinality not greater than Weight of Space and Unions Equal

Theorem
Let $T$ be a topological space.

Let $\FF$ be a set of open sets of $T$.

There exists a subset $\GG \subseteq \FF$ such that:
 * $\ds \bigcup \GG = \bigcup \FF$

and:
 * $\card \GG \le \map w T$

where:
 * $\map w T$ denotes the weight of $T$
 * $\card \GG$ denotes the cardinality of $\GG$.

Proof
By definition of weight of $T$ there exists a basis $\BB$ of $T$ such that:
 * $(1): \quad \card \BB = \map w T$

Let:
 * $\BB_1 = \set {W \in \BB: \exists U \in \FF: W \subseteq U}$

By definition of subset:
 * $\BB_1 \subseteq \BB$

Then by Subset implies Cardinal Inequality:
 * $(2): \quad \card {\BB_1} \le \card \BB$

By definition of set $\BB_1$:
 * $\forall W \in \BB_1: \exists U \in \FF: W \subseteq U$

Then by the Axiom of Choice there exists a mapping $f$ from $\BB_1$ into $\FF$ such that
 * $(3): \quad \forall U \in \BB_1: U \subseteq f \sqbrk U$

Let:
 * $\GG = \Img f$

where $\Img f$ denotes the image of $f$.

By hypothesis:
 * $\GG \subseteq \FF$

Then by Union of Subset of Family is Subset of Union of Family:
 * $\ds \bigcup \GG \subseteq \bigcup \FF$

By definition of set equality, to prove $\ds \bigcup \GG = \bigcup \FF$ it is sufficient to show:
 * $\ds \bigcup \FF \subseteq \bigcup \GG$

Let $\ds x \in \bigcup \FF$.

By definition of union there exists a set $A$ such that:
 * $A \in \FF$ and $x \in A$

Because $A$ is open, then by definition of basis there exists $U \in \BB$ such that:
 * $x \in U \subseteq A$

By definition of the set $\BB_1$:
 * $U \in \BB_1$

Because $\GG = \Img f$:
 * $f \sqbrk U \in \GG$

By $(3)$ and Set is Subset of Union:
 * $\ds U \subseteq f \sqbrk U \subseteq \bigcup \GG$

Thus by definition of subset:
 * $\ds x \in \bigcup \GG$

This ends the proof of inclusion.

By Cardinality of Image of Mapping not greater than Cardinality of Domain:
 * $\card {\Img f} \le \card {\BB_1}$

Thus by $(1)$ and $(2)$:
 * $\card \GG \le \map w T$