Okay, so this is lemma 2.11 in chapter 4 of Mendelson's Introduction to Topology (p. 117):
Let $\{{X_\alpha\}}_{\alpha \in I}$ be an indexed family of topological spaces, each of which is connected. Let $x$ and $x'$ be two points of $ = \prod_{\alpha \in I}X_\alpha$ such that $x_\alpha = x'_\alpha$ except on a finite set of indices $I' \subset I$ and let $f:X \rightarrow \{0,1\}$ be continuous, then $f(x) = f(x')$.
This seems to be related to a statement like this:
$R^n$ can be embedded in $R^{k}$ where $n$, $k$ $\in \mathbb N$, $k > n > 0$.
To my mind, the finite set of indices $I'$ would be analogous to the $n$ in $R^n$, with $I$ being analogous to $k$.
Is this a correct interpretation?
The points $x$ and $x'$ are both in the same connected set $\prod_{\alpha \in I} C_\alpha$, with $C_\alpha = X_\alpha$ for $\alpha \in I'$ and $C_\alpha = \{x_\alpha\}=\{x'_\alpha\}$ for $\alpha \notin I'$, which is connected, being homeomorphic to $\prod_{\alpha \in I} X_\alpha$ via the obvious restriction/projection map. Hence $f$ (mapping to $\{0,1\}$ and being continuous) is constant on that set.
This is assuming that you've already shown finite products of connected spaces are connected (which is a first step towards arbitrary products, our target here).