I know that the product topology on $\prod_{i=1}^\infty\mathbb{R}$ has a basis of the form:
$$ \prod_{i=1}^N U_i \times \prod_{i=N+1}^\infty \mathbb{R} \tag{*} $$
where $U_i\subseteq \mathbb{R}$ is open in $\mathbb{R}$. I'm trying to give an example of an open set not of that form, and just wanted to verify the following example:
Let $U_N\in \prod_{i=1}^\infty\mathbb{R}$ given by
$$ U_N:= \prod_{i=1}^N (N,N+1) \times \prod_{i=N+1}^\infty \mathbb{R} \quad. $$
Given $\overline{a}=(a_1,a_2,a_3,...)$, if $\overline{a}\in U_i$ then $a_1\in(i,i+1)$, and if $\overline{a}\in U_{i+1}$ then $a_1\in(i+1,i+2)$. Hence $U_i\cap U_{i+1}=\emptyset.$
I think that $U=\underset{N=1}{\overset{\infty}{\cup} }U_N $ is not of the form (*), but is open.
Since I have a tendency to get confused I was hoping if someone could verify this is indeed the case, and perhaps point me to a more exotic example which I can't see right now.
Your example is fine, although you are trying too hard.
You can start with the following question: "what open subset of $\mathbb{R}^2$ is not of the form $I_1\times I_2$ for some open subsets $I_1,I_2\subseteq\mathbb{R}$?" And then simply generalize it to $\prod\mathbb{R}$. Like for example two overlapping squares forming a non-square:
$$U_1=(0,2)\times (0,2)\times \prod \mathbb{R}$$ $$U_2=(1,3)\times (1,3)\times \prod \mathbb{R}$$ $$U=U_1\cup U_2$$