If $\mathbb{R}^\ast$ is the one point compactification of $\mathbb{R}$, $A$ is an infinite set, is it true that if $U$ is an open set in the product topology $(\mathbb{R}^\ast)^A$ such that $U\supset\mathbb{R}^A$, then $U$ contains $p_{a_1}^{-1}(\mathbb{R})\cap...\cap p_{a_n}^{-1}(\mathbb{R})$ for some $a_1,...,a_n\in A$, where $p_{a_i}$ is the $a_i$-th coordinate map from $(\mathbb{R}^\ast)^A$ to $\mathbb{R}^\ast$.
This is something I think of while trying to understand the proof of Kolmogorov's extension theorem given in Folland's real analysis on page 328.
No. For instance, let $A=\mathbb{N}$ and consider the sequence $(x_n)$ in $(\mathbb{R}^*)^{\mathbb{N}}$ where the first $n$ coordinates of $x_n$ are $n$ and the rest are $\infty$. Note that $(x_n)$ converges to the point $x$ whose coordinates are all $\infty$, and so the set $C=\{x_n:n\in\mathbb{N}\}\cup\{x\}$ is closed in $(\mathbb{R}^*)^{\mathbb{N}}$. Its complement is an open set that contains $\mathbb{R}^{\mathbb{N}}$, but does not contain any set of the form $p_{a_1}^{-1}(\mathbb{R})\cap\dots\cap p_{a_n}^{-1}(\mathbb{R})$ since $x_N$ is in that set if $N$ is greater than all of $a_1,\dots,a_n$.