Product Space is T0 iff Factor Spaces are T0

Theorem
Let $T_A = \left({S_A, \tau_A}\right)$ and $T_B = \left({S_B, \tau_B}\right)$ be topological spaces.

Let $T = T_A \times T_B$ be the product space formed from $T_A$ and $T_B$.

Then $T$ is a $T_0$ (Kolmogorov) space iff $T_A$ and $T_B$ are themselves both $T_0$ (Kolmogorov) spaces.

General Result
Let $\mathbb S = \left \langle {\left({S_\alpha, \tau_\alpha}\right)}\right \rangle_{\alpha \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.

Let $\displaystyle T = \left({S, \tau}\right) = \prod \left({S_\alpha, \tau_\alpha}\right)$ be the product space of $\mathbb S$.

Then $T$ is a $T_0$ (Kolmogorov) space iff each of $\left({S_\alpha, \tau_\alpha}\right)$ is a $T_0$ (Kolmogorov) space.

Proof
Let $T$ be a $T_0$ (Kolmogorov) space.

From:
 * Subspace of Product Space Homeomorphic to Factor Space
 * Separation Properties Preserved in Subspace
 * Separation Axioms Preserved under Homeomorphism

it follows that $T_A$ and $T_B$ are both $T_0$.

Now suppose that $T$ is not $T_0$.

Then $\exists p, q \in S_A \times S_B: p \ne q$ and $\forall U \in \tau: \{p,q\} \subseteq U$ or $\{p,q\}\cap U = \varnothing$.

If $p \ne q$, then they are different in at least one coordinate.

WLOG, let $p_A \ne q_A$.

Let $V \in \tau_A$ such that $p_A \in V$.

Then from Projection from Product Topology is Continuous $\operatorname{pr}_A^{-1}(V) \in \tau$.

Since $p \in \operatorname{pr}_A^{-1}(V)$, we have that $q\in \operatorname{pr}_A^{-1}(V)$.

From Projections are Surjections, $\operatorname{pr}_i(q) = q_A \in V$.

So $\forall V\in\tau_A: \{p_A, q_A\} \subseteq V$ or $\{p_A, q_A\}\cap V = \varnothing$.

Thus, by definition, $T_A$ is not $T_0$.

Proof of General Result
Suppose $\exists \beta: \left({S_\beta, \tau_\beta}\right)$ is not a $T_0$ space.

Then $\exists a, b \in S_\beta$ such that $\forall U_\beta\in \tau_\beta$, either $a, b \in U_\beta$ or $a, b \notin U_\beta$.

Consider the elements $y, z \in S$ defined as:
 * $y = \left \langle {x_\alpha}\right \rangle: x_\alpha = \begin{cases}

s_\alpha & : \alpha \ne \beta \\ a & : \alpha = \beta \end{cases}$


 * $z = \left \langle {x_\alpha}\right \rangle: x_\alpha = \begin{cases}

s_\alpha & : \alpha \ne \beta \\ b & : \alpha = \beta \end{cases}$

That is, $y$ and $z$ match (arbitrarily) on all ordinates except that for $\beta$.

Let $H \subseteq S: y \in H$.

Then $z \in H$ as $\forall U_\beta \in \operatorname{pr}_\beta \left({H}\right): b \in U_\beta$

Similarly, let $K \subseteq S: z \in K$.

Then $y \in K$ as $\forall U_\beta \in \operatorname{pr}_\beta \left({H}\right): a \in U_\beta$

So $T$ is not a $T_0$ (Kolmogorov) space.

Suppose $T$ is not a $T_0$ (Kolmogorov) space.

Then $\exists a, b \in S, a \ne b$ such that for all $U \in \tau$, either $a, b \in U$ or $a, b \notin U$.

Then $a$ and $b$ are different in at least one ordinate.

Suppose, $a_m = p, b_m = q$ for some ordinate $m$.

Then either $a_m, b_m \in U_m$ or $a_m, b_m \notin U_m$.

It follows that $\left({S_m, \tau_m}\right)$ is not a $T_0$ (Kolmogorov) space.