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 = Disjoint Open Sets remain Disjoint with one 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 $\displaystyle U'_n = U_n \setminus \bigcup_{i \mathop \le n} V_i^-$ and $\displaystyle V'_n = V_n \setminus \bigcup_{i \mathop \le n} U_i^-$. Define $\displaystyle U' = \bigcup_{n \mathop \in \N} U'_n$ and $\displaystyle V' = \bigcup_{n \mathop \in \N} V'_n$. 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: :$\displaystyle \bigcup_{i \mathop \le n} V_i^-$ is closed. By Open Set minus Closed Set is Open: :$\displaystyle U'_n = U_n \setminus \bigcup_{i \mathop \le n} V_i^-$ is open. By {{Defof|Topological Space}}: :$\displaystyle U' = \bigcup_{n \mathop \in \N} U'_n$ 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 }} {{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}}