Product Space is Completely Hausdorff iff Factor Spaces are Completely Hausdorff/Necessary Condition

Theorem
Let $\mathbb S = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces for $\alpha$ in some indexing set $I$.

Let $\displaystyle T = \struct {S, \tau} = \displaystyle \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\mathbb S$.

Let $T$ be a completely Hausdorff space.

Then for each $\alpha \in I$, $\struct {S_\alpha, \tau_\alpha}$ is a completely Hausdorff space.

Proof
Suppose $T$ is a completely Hausdorff space.

As $S_\alpha \ne \O$ we also have $S \ne \O$ by the axiom of choice.

Let $\alpha \in I$.

From Subspace of Product Space is Homeomorphic to Factor Space, $\struct {S_\alpha, \tau_\alpha}$ is homeomorphic to a subspace $T_\alpha$ of $T$.

From completely Hausdorff property is hereditary, $T_\alpha$ is completely Hausdorff.

From Completely Hausdorff Space is Preserved under Homeomorphism, $\struct {S_\alpha, \tau_\alpha}$ is completely Hausdorff.

As $\alpha \in I$ was arbitrary, the result follows.