Theorem: Second-Countable T3 Space is T5

Let $T = \struct {S, \tau}$ be a $T_3$ space which is also second-countable.
Then $T$ is a $T_5$ space.

Proof:
Let $A, B \subseteq S$ with $A^- \cap B = A \cap B^- = \O$.
For each $x \in A$, since $T$ is $T_3$:
:$\exists P, Q \in \tau: x \in P, B^- \subseteq Q, P \cap Q = \O$
Let $\BB$ be a basis for $T$.
Then:
:$\exists U \in \BB: x \in U \subseteq P$
Notice that:
{{begin-eqn}}
{{eqn | o =
      | r = U^- \cap B
}}
{{eqn | o = \subseteq
      | r = U^- \cap B^-
      | c = Set Intersection Preserves Subsets; Set is Subset of its Topological Closure
}}
{{eqn | o = \subseteq
      | r = P^- \cap Q
      | c = Set Intersection Preserves Subsets; Topological Closure of Subset is Subset of Topological Closure
}}
{{eqn | r = \O
      | c = Open Set Disjoint from Set is Disjoint from Closure
}}
{{end-eqn}}
By Subset of Empty Set, $U^-$ and $B$ are disjoint.
Since $T$ is second-countable, $\BB$ is countable.
Doing the above process for each $x \in A$ yields a subset $\set {U_n}_{n \mathop \in \N}$ of $\BB$.
Doing a similar process for each $y \in B$ yields another subset $\set {V_n}_{n \mathop \in \N}$ of $\BB$.
These sets are open sets by definition.
Define $\ds U'_n = U_n \setminus \bigcup_{i \mathop \le n} V_i^-$ and $\ds V'_n = V_n \setminus \bigcup_{i \mathop \le n} U_i^-$.
Define $\ds U' = \bigcup_{n \mathop \in \N} U'_n$ and $\ds V' = \bigcup_{n \mathop \in \N} V'_n$.
We show that $U'$ and $V'$ are disjoint open sets containing $A$ and $B$ respectively.
For any $n \in \N$, we have that $U_n$ is open.
From Topological Closure is Closed:
:$V_i^-$ is closed for each $i \le n$.
From Finite Union of Closed Sets is Closed in Topological Space:
:$\ds \bigcup_{i \mathop \le n} V_i^-$ is closed.
By Open Set minus Closed Set is Open:
:$\ds U'_n = U_n \setminus \bigcup_{i \mathop \le n} V_i^-$ is open.
By {{Defof|Topological Space}}:
:$\ds U' = \bigcup_{n \mathop \in \N} U'_n$ is open.
Similarly, $V'$ is open.
Let $y \in B$.
By construction, there is some $k \in \N$ where $y \in V_k$.
From above we see that $U_i^-$ and $B$ are disjoint for all $i \in \N$.
So $y \notin U_i^-$ for every $i \le k$.
Hence $y \in V_k \setminus \bigcup_{i \mathop \le k} U_i^- = V'_k \subseteq V'$.
By {{Defof|Subset}}, $B \subseteq V'$.
Similarly, $A \subseteq U'$.
Let $i, j \in \N$.
We show that $U'_i, V'_j$ are disjoint.
{{WLOG}} suppose $i \le j$.
Then:
{{begin-eqn}}
{{eqn | l = U'_i
      | r = U_i \setminus \bigcup_{k \mathop \le i} V_k^-
}}
{{eqn | o = \subseteq
      | r = U_i
      | c = Set Difference is Subset
}}
{{eqn | o = \subseteq
      | r = U_i^-
      | c = Set is Subset of its Topological Closure
}}
{{eqn | o = \subseteq
      | r = \bigcup_{k \mathop \le j} U_k^-
      | c = Set is Subset of Union of Family
}}
{{eqn | ll = \leadsto
      | l = U'_i \cap V'_j
      | r = \O
      | c = Empty Intersection iff Subset of Relative Complement
}}
{{end-eqn}}
Now:
{{begin-eqn}}
{{eqn | l = U' \cap V'
      | r = \paren {\bigcup_{i \mathop \in \N} U'_i} \cap \paren {\bigcup_{j \mathop \in \N} V'_j}
}}
{{eqn | r = \bigcup_{\tuple {i, j} \mathop \in \N \times \N} \paren {U'_i \cap V'_j}
      | c = Intersection Distributes over Union
}}
{{eqn | r = \bigcup_{\tuple {i, j} \mathop \in \N \times \N} \O
      | c = 
}}
{{eqn | r = \O
      | c = Union is Empty iff Sets are Empty
}}
{{end-eqn}}
Hence $U'$ and $V'$ are disjoint.
Since $A, B$ are arbitrary, $T$ is a $T_5$ space.
{{qed}}