T3 Lindelöf Space is T4 Space/Proof 1

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

Lemma 1
Let $\UU = \set{U \in \tau : U^- \cap B = \O}$.

From Lemma 1:
 * $\forall a \in A : \exists U_a \in \tau: a \in U_a : U_a^- \cap B = \O$

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

From Closed Subspace of Lindelöf Space is Lindelöf Space:
 * $\struct{A, \tau_A}$ is a Lindelöf subspace

where $\tau_A$ is the subspace topology on $A$.

By definition of Lindelöf space:
 * there exists a countable subcover $\family{U_n}_{n \in \N}$ of $\UU$ for $A$

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

Similar to $A$ above, there exists a countable subcover $\family{V_n}_{n \in \N}$ of $\VV$ for $B$.

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

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

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

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

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

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

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

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

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

Lemma 3
We have:

Similarly:

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