Finite Product Space is Connected iff Factors are Connected/General Case

Theorem
Let $I$ be an indexing set.

Let $\family {T_\alpha}_{\alpha \mathop \in I}$ be an indexed family of topological spaces.

Let $T = \ds \prod_{\alpha \mathop \in I} T_\alpha$ be the Cartesian space of $\family {T_\alpha}_{\alpha \mathop \in I}$.

Let $T = \ds \overline {\bigcup_{\alpha \mathop \in I} S_\alpha}$.

Let $\tau$ be a topology on $T$ such that the subsets ${S'}_\alpha \subseteq \ds \prod T_\alpha$ where ${S'}_\alpha = \set {\family {y_\beta} \in T: y_\beta = x \beta \text { for all } \beta \ge \alpha}$ is homeomorphic to $S_{\alpha - 1} \times T_\alpha$.

Then $T$ is connected each of $T_\alpha: \alpha \in I$ are connected.

Proof
Let the Axiom of Choice be assumed.

Let $I$ be well-ordered.

Let $x = \family {x_\alpha} \in T$ be some arbitrary fixed element of $T$.

Let $S_\alpha = \set {\family {y_\beta} \in T: y_\beta = x \beta \text { for all } \beta \ge \alpha}$.

We have that $S_\alpha$ is homeomorphic to $S_{\alpha - 1} \times T_\alpha$.

Then from Finite Product Space is Connected iff Factors are Connected, $S_\alpha$ is connected $S_{\alpha - 1}$ is.

Let $\alpha$ be a limit ordinal.

Then:
 * $S_\alpha = \ds \paren {\bigcup_{\beta \mathop < \alpha} S_\beta}^-$

where $X^-$ denotes the closure of $X$.

So if each $S_\beta$ is connected for $\beta < \alpha$, it follows that $S_\alpha$ must likewise be connected, as $\family \gamma$ is nested.

Thus:
 * $T = \ds \overline {\bigcup_{\alpha \mathop \in I} S_\alpha}$ is connected.