Let $D = \prod\nolimits_{j=1}^\infty \{0, 1\}$ be an infinite product in the product topology. Show that the eventually constant elements of $D$ are dense in $D$.
I have a hard time wrapping my head around what exactly it means to be dense. I know that the definition is a subset $A$ of a space $X$ where $\bar{A} = X$ but I have no idea how to use that here.
A set $A \subseteq X$ is dense iff for every non-empty open subset $O$ of $X$ we have that $ O \cap A \neq \emptyset$.
If $A$ satisfies this condition, let $C$ be a closed subset such that $A \subseteq C$. The open set $X\setminus C$ cannot be non-empty, as it is disjoint from $A$, and so $C=X$, and so the only closed set containing $A$ is $X$, hence $\overline{A} = X$. The reverse implication is similar.
Now if $O$ is non-empty and open in $X =\prod_{j=1}^{\infty}\{0,1\} = \{0,1\}^\mathbb{N}$ then by the definition of he product topology it contains a non-empty basic open subset $\prod_i O_i$ where $O_i = \{0,1\}$ except that $O_i$ is a proper, non-empty open subset of $\{0,1\}$, so of the form $\{d_i\}$, for $i \in F$ where $F \subseteq \mathbb{N}$ is finite.
Then the sequence $s_n$ that is $d_n$ for $n \in F$ and $d_n = 0$ for all other $n$ is eventually constant and lies in this basic open set, hence in $A$. So $A \cap O \neq \emptyset$, as required.
We could also have chosen $d_n=1$ for $n \notin F$, and see that both the eventually-$0$ as well as the eventually-$1$ sequences are dense in $X$.