Can a continuous real-valued function on a large product space depend on uncountably many coordinates? Is there a reasonably well-behaved topological space $X$ (ideally Polish), a set $\kappa$, and a continuous function $g: X^\kappa\to\mathbb{R}$ that depends on uncountable many coordinates?
If $X$ is a compact Hausdorff space, the answer is known to be no. To see this, note that the family of all continuous functions depending on only finitely many coordinates satisfies the conditions for the Stone-Weierstrass theorem and is, therefore, uniformly dense. The argument can be found in textbooks.
 A: Bockstein's theorem
Bockstein, M., Un théorème de séparabilité pour les produits topologiques, Fundam. Math. 35, 242-246 (1948). ZBL0032.19103.
This is the case of a product $\prod_{t \in T} X_t$ where all factors are second-countable.  I that case any continuous function
$\prod_{t \in T} X_t \to \mathbb R$ depends on countably many coordinates.
PLUG... See Theorem 2.1 in
Edgar, G. A., Measurability in a Banach space, Indiana Univ. Math. J. 26, 663-677 (1977). ZBL0361.46017.
where the special case $X = \mathbb R$ is done.  That is, a continuous function $\mathbb R^T \to \mathbb R$ depends on only countably many coordinates.
A: Let $X$ be an uncountable discrete space with a distinguished element $0$.  We view the product space $X^X$ as the space of maps $\phi: X \to X$.  The set
$$ E := \{ \phi \in X^X: \phi(\phi(0)) = 0 \}$$
is easily seen to be clopen, hence the indicator function $1_E: X \to {\bf R}$ is continuous, but depends on all of the (uncountably many) coordinates of $X^X$.
The key point here (which was inspired by Nate's comment based on the earlier incorrect attempt at solving this problem) is that deciding whether a given map $\phi$ belongs to $E$ requires only a finite number of (adaptive) evaluations of $\phi$, but the set of (non-adaptive) locations where $\phi$ could potentially need to be evaluated is uncountable.
Note that a similar construction works for $X \times \{0,1\}^X$ using the set $E := \{ (x, \phi) \in X \times \{0,1\}^X: \phi(x)=0\}$; thus even a single highly non-compact factor is enough to generate a counterexample. (But I am not sure what happens if one insists that all of the factors be sigma-compact, in particular can one construct a continuous function $f: {\bf N}^{\bf R} \to {\bf R}$ that depends on uncountably many coordinates?)
