T3 Space with Sigma-Locally Finite Basis is T4 Space/Lemma 1

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

Let $\BB$ be a $\sigma$-locally finite basis. Let $F$ be a closed subset of $T$.

Let $X \subseteq S$ be disjoint from $F$.

Then there exists a countable open cover $\WW = \set{W_n : n \in \N}$ of $X$:
 * $\forall n \in \N : W_n^- \cap F = \O$

Proof
Let $\UU = \set {U \in \BB : U^- \cap F = \O}$.

Lemma 2
From Lemma 2:
 * $\forall x \in X : \exists U_x \in \BB : x \in U_x : U_x^- \cap B = \O$

Hence:
 * $\forall x \in X : \exists U_x \in \UU$

By definition of open cover:
 * $\UU$ is an open cover of $X$

By definition of $\sigma$-locally finite:
 * $\BB = \ds \bigcup_{n \in \N} \BB_n$

where $\BB_n$ is a locally finite set of subsets.

For each $n \in \N$, let:
 * $\UU_n = \UU \cap \BB_n$

From User:Leigh.Samphier/Topology/Subset of Locally Finite Set of Subsets is Locally Finite:
 * $\UU_n$ is locally finite

For each $n \in \N$, let:
 * $W_n = \ds \bigcup \set{W : W \in \UU_n}$

From :
 * $W_n \in \tau$

From User:Leigh.Samphier/Topology/Union of Closures of Elements of Locally Finite Set is Closed:
 * $W_n^- = \ds \bigcup \set{W^- : W \in \UU_n}$

We have:

Let $x \in X$.

From Lemma 2:
 * $\exists U_x \in \BB : x \in U_x : U_x^- \cap F = \O$

By definition of $\UU$:
 * $U_x \in \UU'$

As $\BB = \ds \bigcup_{n \in \N} \BB_n$:
 * $\exists n \in \N : U_x \in \BB_n : x \in U_x : U_x^- \cap F = \O$

Hence:
 * $U_x \in \UU_n$

From Set is Subset of Union:
 * $U_x \subseteq \ds \bigcup \UU_n = W_n$

Hence:
 * $x \in W_n$

By definition, $\WW = \set{W_n : n \in \N}$ is a countable open cover of $X$.