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

Proof
Let $A$ and $B$ be disjoint closed subsets of $T$.

Existence of Countable Cover of $A$
Let $\UU = \set {U \in \BB : U^- \cap B = \O}$.

From Characterization of T3 Space:
 * $\forall a \in A : \exists W_a \in \tau: a \in W_a : W_a^- \cap B = \O$

By definition of basis:
 * $\exists B_a \in \BB : a \in B_a \subseteq W_a$

From Topological Closure of Subset is Subset of Topological Closure:
 * $B_a^- \subseteq W_a^-$

From Set Intersection Preserves Subsets:
 * $B_a^- \cap B = \O$

Hence:
 * $B_a \in \UU$

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

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 :
 * $\UU_n$ is locally finite

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

From :
 * $U_n^- = \ds \bigcup \set{W^- : W \in \UU_n}$

From :
 * $U_n^- = \ds \bigcup \set{W^- : W \in \UU_n}$

Hence:
 * $U_n^- \cap B = \O$

Let $\VV = \set {V \in \BB : V^- \cap A = \O}$.

Similarly to above:
 * $\VV$ is an open cover of $B$

By definition of cover:
 * $\ds A \subseteq \bigcup_{n \mathop \in \N} U_n$

We have:
 * $\ds A \cap \paren{\bigcup_{n \mathop \in \N} V_n^-} = \O$

From Subset of Set Difference iff Disjoint Set:
 * $(1) \quad \ds A \subseteq \paren {\bigcup_{n \mathop \in \N} U_n } \setminus \paren {\bigcup_{n \mathop \in \N} V_n^-}$

Similarly:
 * $(2) \quad \ds B \subseteq \paren {\bigcup_{n \mathop \in \N} V_n} \setminus \paren {\bigcup_{n \mathop \in \N} U_n^-}$

For each $n \in \N$, let:
 * $U'_n = U_n \setminus \paren {\ds \bigcup_{p \mathop \le n} V_p^-}$

For each $n \in \N$, let:
 * $V'_n = V_n \setminus \paren {\ds \bigcup_{p \mathop \le n} U_p^-}$

Lemma 2
Let:
 * $U = \ds \bigcup_{n \mathop \in \N} U'_n$

and
 * $V = \ds \bigcup_{n \mathop \in \N} V'_n$

From Lemma 2:
 * $U \cap V = \O$

Lemma 3
We have:

Similarly, from $(2)$ abvove:

It has been shown that there exists $U, V \in \tau$ such that $A \subseteq U, B \subseteq V$ and $U \cap V = \O$.

Hence, by definition, $T$ is a $T_4$ space.