Since $S_\alpha \ne \O$ we also have $S \ne \O$ by the axiom of choice. Let $z \in S$. From Subspace of Product Space Homeomorphic to Factor Space, every $\struct {S_\alpha, \tau_\alpha}$ is homeomorphic to the subspace $T_\alpha$ of $T$ defined by: :$T_\alpha = \set {x \in S: \forall \beta \in I \setminus \set \alpha: x_\beta = z_\beta}$ and the homeomorphism $h : S_\alpha \to T_\alpha$ is given by: :$\forall x_\alpha \in S_\alpha : \map h {x_\alpha} = x : \forall \beta \in I : x_\beta = \begin {cases} x_\alpha & \beta = \alpha \\ z_\beta & \beta \ne \alpha \end {cases}$  From Inclusion Mapping is Continuous, the inclusion mapping $i: T_\alpha \to S$ is continuous. From Continuity of Composite Mapping, the composite mapping $i \circ h : S_\alpha \to S$ is continuous. Let $\pr_\alpha: S \to S_\alpha$ be the projection of $S$ to $S_\alpha$. For all $x_\alpha \in S_\alpha$: :$\map {\pr_\alpha} {\map {i \circ h} {x_\alpha}} = x_\alpha$  Let $A, B \subseteq S_\alpha$ be an arbitrary pair of disjoint closed sets in $\struct {S_\alpha, \tau_\alpha}$. Then: {{begin-eqn}} {{eqn | l = \map {\pr_\alpha^\gets} A \cap \map {\pr_\alpha^\gets} B       | r = \map {\pr_\alpha^\gets} {A \cap B}       | c = Preimage of Intersection under Mapping }} {{eqn | r = \map {\pr_\alpha^\gets} \O }} {{eqn | r = \O }} {{end-eqn}}  From Projection from Product Topology is Open and Continuous:General Result :$\pr_\alpha: S \to S_\alpha$ is continuous. From Continuity Defined from Closed Sets: :$\map {\pr_\alpha^\gets} A, \map {\pr_\alpha^\gets} B$ are disjoint closed sets in $T$   By definition of $T_4$ space: :$\exists U, V \in \tau : \map {\pr_\alpha^\gets} A \subseteq U, \map {\pr_\alpha^\gets} B \subseteq V : U \cap V = \O$ By definition of continuity: :$\map {\paren {i \circ h}^\gets} U, \map {\paren{i \circ h}^\gets} V$ are open sets in $S_\alpha$. From Preimage of Intersection under Mapping: :$\map {\paren {i \circ h}^\gets} U, \map {\paren{i \circ h}^\gets} V$ are disjoint open sets in $S_\alpha$. From Preimage of Subset is Subset of Preimage: :$\map {\paren {i \circ h}^\gets} {\map {\pr_\alpha^\gets} A} \subseteq \map {\paren{i \circ h}^\gets} U$ :$\map {\paren {i \circ h}^\gets} {\map {\pr_\alpha^\gets} B} \subseteq \map {\paren{i \circ h}^\gets} V$ Now: {{begin-eqn}} {{eqn | o =        | r = x_\alpha \in \map {\paren{i \circ h}^\gets} {\map {\pr_\alpha^\gets} A} }} {{eqn | ll = \leadstoandfrom       | o =        | r = \map {\paren{i \circ h} } {x_\alpha} \in \map {\pr_\alpha^\gets} A       | c = {{Defof|Inverse Image Mapping}} }} {{eqn | ll = \leadstoandfrom       | o =        | r = \map {\pr_\alpha} {\map {i \circ h} {x_\alpha} } \in A       | c = {{Defof|Inverse Image Mapping}} }} {{eqn | ll = \leadstoandfrom       | o =        | r = x_\alpha \in A }} {{end-eqn}} From the definition of set equality: :$\map {\paren{i \circ h}^\gets} {\map {\pr_\alpha^\gets} A} = A$ Similarly: :$\map {\paren{i \circ h}^\gets} {\map {\pr_\alpha^\gets} B} = B$ It follows that $A$ and $B$ are contained in the disjoint open sets $\map {\paren {i \circ h}^\gets} U, \map {\paren {i \circ h}^\gets} V$, respectively. Since $A$ and $B$ were an arbitrary pair of disjoint closed sets in $\struct {S_\alpha, \tau_\alpha}$ then $\struct {S_\alpha, \tau_\alpha}$ is a $T_4$ space. {{qed}} {{AoC}}