I'll quote the beginning of the example from the document you have linked to.
Example 3: Let $X$ be an uncountable set and let $\{0,1\}$ have the discrete topology. Consider $\mathcal P(X) = \{0,1\}^X$ with the product topology. Let $\mathcal A\subseteq \mathcal P(X)$ be the collection of all uncountable subsets of $X$. $\mathcal A$ is
not open; indeed every basic open contains finite sets. However, we claim that $\mathcal A$ is sequentially open.
The important point is notice that the author identifies $\{0,1\}^X$ with $\mathcal P(X)$. Indeed, there is a very natural bijection between these two sets (and it occurs frequently in mathematics, which might be why it is mentioned in the text without any details).
For every subset $S\subseteq X$ you have the corresponding function $\chi_S \in \{0,1\}^X$ which maps elements of $S$ to $1$ and other elements to $0$
$$\chi_S(x)=
\begin{cases}
1 & x\in S, \\
0 & \text{otherwise}.
\end{cases}
$$
The assignment $S\mapsto \chi_S$ is a bijection between $\mathcal P(X)$ and $\{0,1\}^X$.
Now if we look at the space $\{0,1\}^X$ then the basic open sets are obtained in this way: Choose finitely many coordinates $k_1,\dots,k_n$ and choose values $v_1,\dots,v_n\in\{0,1\}$ which you want to prescribe for these coordinates. Then you get a basic neighborhood consisting of all functions with these prescribed values, i.e. $f(k_j)=v_j$.
What happens if we transfer this topology (using the bijection we mentioned above) to the set $\mathcal P(X)$? Well, for every basic neighborhood we have fixed some finite set $F$ of indices where the functions have to attain the value $1$. If this function is of the form $\chi_S$, this is the same as saying $F\subseteq S$. We also prescribed some finite sets $G$ such that for $x\in G$ we want $\chi_S(x)=0$. This is the same as requiring $S\cap G=\emptyset$.
So the basic open sets in $\mathcal P(X)$ are like this: For any finite sets $F,G\subseteq X$ you get a basic set
$$\mathcal O_{F,G}=\{S\subseteq X; F\subseteq S, G\cap S=\emptyset\}.$$
You can notice, that for any choice of $F$, $G$ such that $F\cap G=\emptyset$, this basic set contains at least one finite set. (For $F\cap G\ne\emptyset$, the set $\mathcal O_{F,G}$ described here is empty.)
$(1).$ Unless some $U_i$ is empty, the set $U=\prod_{i\in \Bbb N}U_i$ is not open in the product topology:
Suppose $U$ is open and $p\in U.$ Then $p\in C\subset U$ for some basic open set $C=\prod_{i\in \Bbb N}C_i$ where each $C_i$ is non-empty & open in $X_i$ and the set $D=\{i:C_i\ne X_i\}$ is finite.
The key is that D is not empty. Consider, for some (any) $j\in D,$ the projection $p_j:\prod_{i\in \Bbb N} X_i\to X_j$ where $p_j(x_1, x_2, x_3,...)=x_j.$ The image $p_j(C)$ of $C$ under $p_j$ is a subset of the image $p_j(U)$ because $C\subset U.$ But then $X_j=p_j(C)\subset p_j(U)=U_j,$ which is false.
In other words: If $U$ is not empty then for any $j,$ the set of the $j$-th co-ordinates of the members of $U$ is $U_j,$ but if $U$ is open and non-empty then for any $j\in D,$ the set of the $j$-th co-ordinates of the members of $U$ is $X_j.$
$(2).$ It is possible that the box topology is compact. For example if $X_i$ is the only non-empty open subset of each $X_i$ then the only non-empty open subset of $X=\prod_{i\in \Bbb N}X_i$ in the box topology is $X$ itself.
On the other hand, suppose each $X_i$ is a two-point discrete space (which is obviously compact). Then the box topology on $\prod_{i\in \Bbb N}X_i$ is an infinite discrete space (which is obviously not compact).
Best Answer
The open unit circle in $\mathbb{R}^2$ is an example, or $\{(x,y): x \neq y\}$. The plane has the product topology wrt the usual topology on $\mathbb{R}$. These sets are open but not Cartesian products of two sets.
The product of two finite topologies is always finite: it has a finite base, and we can only form finitely many unions from it.
It surely is uncountable. Prove it. The standard base is countable, but in most cases there will be uncountably many different unions formed from it.
It can and often will be uncountable, as in the case of the Cantor cube $\{0,1\}^\mathbb{N}$. e.g.