T3 Space with Sigma-Locally Finite Basis is Perfectly T4 Space/Lemma 2

Theorem
Let $T = \struct {S, \tau}$ be a $T_3$ topological space.

Let $\BB$ be a basis for $T$.

Let $G$ be open in $T$.

Let:
 * $\CC = \set{B \in \BB : B^- \subseteq G}$

where $B^-$ denotes the closure of $B$ in $T$.

Then:
 * $\CC$ is a is a cover of $G$

Proof
Let $x \in G$.

From Characterization of T3 Space:
 * $\exists U \in \tau : x \in U : U^- \subseteq G$

where $U^-$ denotes the closure of $U$ in $T$.

By definition of basis:
 * $\exists B \in \BB : x \in B : B \subseteq U$

From Topological Closure of Subset is Subset of Topological Closure:
 * $B^- \subseteq U^-$

From Subset Relation is Transitive:
 * $B^- \subseteq G$

Hence:
 * $B \in \CC$

Since $x$ was arbitrary, we have:
 * $\forall x \in G : C \in \CC : x \in C$

Hence $\CC$ is a cover of $G$ by definition.