I want to rid myself of any misunderstandings I have, so feel free to nitpick my attempt all you want so that in the end it's as clear as possible!
Now, this is the exercise:
\begin{array} { l } { \text { Let } \mathbb { R } ^ { \infty } \text { be the subset of } \mathbb { R } ^ { \omega } \text { consisting of all sequences that are "eventually zero," } } \\ { \text { that is, all sequences } \left( x _ { 1 } , x _ { 2 } , \ldots \right) \text { such that } x _ { i } \neq 0 \text { for only finitely many values } } \\ { \text { of } i . \text { What is the closure of } \mathbb { R } ^ { \infty } \text { in } \mathbb { R } ^ { \omega } \text { in the box and product topologies? Justify } } \\ { \text { your answer. } } \end{array}
I will first show that $\overline{\mathbb{R}^{\infty}} = \mathbb{R^\omega}$ in the product topology. Indeed, take any $x \in \mathbb{R^\omega}$ and let $U$ be an open set of $\mathbb{R^\omega}$ such that $x \in U$. Then $U$ contains a basis element $B$ of the product topology such that: $$x \in B = \displaystyle{\prod_{n \in \mathbb{N}} U_{n}} \subset U $$
and $U_n = \mathbb{R} \ \forall n \in \mathbb{N}\setminus F$ where $F$ is a finite subset of $\mathbb{N}$. Now, the sequence $(y_n)_{n \in \mathbb{N}}$ defined by:
$$\begin{align*} &y_n = \pi_n(x) \ \forall n \in F \\ &y_n = 0 \ \forall n \notin F\end{align*}$$
is an element of $B$, by construction. It follows that $(y_n)_{n \in \mathbb{N}} \in B \subset U$, and also by construction we have that $(y_n)_{n \in \mathbb{N}} \in \mathbb{R}^{\infty}$. Then: $$U \bigcap \mathbb{R}^\infty \neq \emptyset$$
and therefore $\overline{\mathbb{R}^{\infty}} = \mathbb{R^\omega}$.
$\\$
Now I will show that $\overline{\mathbb{R}^{\infty}} = \mathbb{R^\infty}$ in the box topology. For this, it suffices to show that any $x \notin \mathbb{R}^{\infty}$ is not in $\overline{\mathbb{R}^{\infty}}$, and this can be done by showing that there exists an open set $U \ni x$ in the product topology such that $U \bigcap \mathbb{R}^{\infty} = \emptyset$. Indeed, by definition if $x = (x_n)_{n \in \mathbb{N}} \notin \mathbb{R}^{\infty}$, then there exists an infinite set $I \subset \mathbb{N}$ such that $x_i \neq 0 \ \forall i \in I$. Now, for: $$U = \displaystyle{\prod_{i \in \mathbb{N}} U_{i} }$$
where $U_i = \mathbb{R} \setminus \{0\} \forall i \in I$ and $U_i = \mathbb{R}$ otherwise. As desired, it's clear that $x \in U$ and $U \bigcap \mathbb{R}^{\infty} = \emptyset$ and we're done.
Have I made any unnecessary or not entirely correct steps here? Is there anything I should make clearer?
EDIT: As I thought, there were some things that could be (and indeed were) improved. Thanks a lot, Brevan and Henno! This kind of thinking is very important and I will always try to bear it in mind so that my proofs are always as clean and clear as possible.
Best Answer
I think it's quite clear and well-explained. Some tips, maybe:
For the first you could just use that $D$ is dense iff it intersects every non-empty open set from a base, which is exactly what you did anyway. The $x$ plays no rôle in the proof. The $(x_n)$ you then construct has no relation to the $x$ you started with. You could also construct $(y_n)$ as $y_n = x_n$ for $n \in F$ and $0$ outside if you want to keep the $x$ (I suppose you want to explicitly show $x$ is in the closure this way). And then $(y_n)$ witnesses $B \cap \mathbb{R}^\infty\neq \emptyset$. Using a different letter is less confusing IMO.
As to the second, just take $\mathbb{R}\setminus\{0\}$ explicitly for those non-zero coordinates, and (like you did) $\mathbb{R}$ for the others. This avoids a minor use of the countable axiom of choice. Be explicit when it's easy to be so. You could also call the non-zero coordinates $I$ as Brevan suggested. You don't need an explicit enumeration. Fewer subscripts is often better.