Basis has Subset Basis of Cardinality equal to Weight of Space

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

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

Then there exists a basis $\mathcal B_0$ of $T$ such that
 * $\mathcal B_0 \subseteq \mathcal B$ and $\left\vert \mathcal B_0 \right\vert = w \left( T \right)$

where
 * $\left\vert \mathcal B_0 \right\vert$ denotes the cardinality of $\mathcal B_0$,
 * $w \left( T \right)$ denotes the weight of $T$.

Proof
Ther are two cases: infinite weight and finite weight.

Case when Weight is Infinite
Let $T$ has infinite weight.

By definition of weight there exists a basis $\mathcal B_1$ of $T$ such that
 * $(1): \quad \left\vert \mathcal B_1 \right\vert = w \left( T \right)$.

We will prove
 * $(2): \quad \forall U \in \mathcal B_1: \exists \mathcal A \subseteq \mathcal B: U = \bigcup \mathcal A \land \left\vert \mathcal A \right\vert \leq w \left( T \right)$.

Let $U \in \mathcal B_1$.

Let $S = \left\{W \in \mathcal B: W \subseteq U \right\}$.

By definition of subset:
 * $S \subseteq \mathcal B$.

By definition of basis: $\bigcup S = U$.

By definition of set $S$, $S$ is set of open subset of $T$.

Then by Existence of Subfamily of Cardinality not greater than Weight of Space and Unions Equal there exists a subset $\mathcal A \subseteq S$ such that
 * $\bigcup \mathcal A = \bigcup S$ and $\left\vert \mathcal A \right\vert \leq w \left( T \right)$.

Thus by Subset Relation is Transitive:
 * $\mathcal A \subseteq \mathcal B$.

Thus $U = \bigcup \mathcal A \land \left\vert \mathcal A \right\vert \leq w \left( T \right)$.

This ends proof of $(2)$.

By $(2)$ and Axiom of Choice there exists a mapping $f: \mathcal B_1 \to \mathcal P \left( \mathcal B \right)$ such that
 * $(3): \quad \forall U \in \mathcal B_1: U = \bigcup f \left( U \right) \land \left\vert f \left( U \right) \right\vert \leq w \left( T \right)$

By Union is Smallest Superset:
 * $\bigcup_{U \in \mathcal B_1} f \left( U \right) \subseteq \mathcal B$.

Set $\mathcal B_0 := \bigcup_{U \in \mathcal B_1} f \left( U \right)$

Now we will show that $\mathcal B_0$ is basis of $T$.

By definition of basis:
 * $\mathcal B \subseteq \tau$.

Thus by Subset Relation is Transitive:
 * $\mathcal B_0 \subseteq \tau$.

Let $A$ be an open subset of $X$.

Let $p$ be a point of $X$ such that $p \in A$.

Then by definition of basis there exists $U \in \mathcal B_1$ such that
 * $p \in U \subseteq A$.

By $(3)$, $U = \bigcup f \left( U \right)$.

Then by definition of union there exists a set $D$ such that
 * $p \in D \in f \left( U \right)$.

By Set is Subset of Union:
 * $D \subseteq U$.

by definition of union:
 * $D \in \mathcal B_0$.

Thus by Subset Relation is Transitive:
 * $\exists D \in \mathcal B_0: p \in D \subseteq A$.

This by definition of basis ends a proof of basis.

By $(1)$ and Cardinality of Image of Mapping not greater than Cardinality of Domain:
 * $\left\vert \operatorname{Im}\left( f \right) \right\vert \leq w \left( T \right)$.

For every $U \in \mathcal B_1$, $\left\vert f \left( U \right) \right\vert \leq w \left( T \right)$.

Then by Cardinality of Union not grateer than Product:
 * $\left\vert \mathcal B_0 \right\vert \leq w \left( T \right) \cdot w \left( T \right)$.

Thus by
 * $\left\vert \mathcal B_0 \right\vert \leq w \left( T \right)$.