I just proved the following theorem:
If $X$ is a regular space that can be written as a countable union of compact subspaces of $X$, then $X$ is paracompact.
I am now working on the following:
Show that $\Bbb{R}^\infty$ is paracompact as a subspace of $\Bbb{R}^\omega$ with the box topology. [Note: $\Bbb{R}^\infty$ consists of all $(x_i)_{i \in \Bbb{N}}$ where $x_i \neq 0$ for only finitely many $i \in \Bbb{N}$]
Here is an attempt of mine. First I want to show a certain function is continuous. Let $k \in \Bbb{N}$ and let $f_k : \Bbb{R}^\omega \to \Bbb{R}^k$ be defined by $f_k(x_1,...,x_k,x_{k+1},...)=(x_1,..,x_k)$. If $U_1 \times ... \times U_k$ is a basis element in $\Bbb{R}^k$, then $f^{-1}_k(U_1 \times ... \times U_k) = U_1 \times ... \times U_k \times \Bbb{R} \times ...$, which is by definition open in the product topology and therefore open in the box topology. When the domain of $f_k$ is restricted to $\Bbb{R} \times ...\times \Bbb{R} \times \{0\} \times ...$, $f_k$ is a homeomorphism from it to $\Bbb{R}^k$. Now, let $A_{n,k} = [-n,n] \times ... \times [-n,n] \times \{0\} ...$ which is compact, being homeomorphic to $[-n,n]^k$ through $f_k$. But clearly $\Bbb{R}^\infty = \bigcup_{(n,k) \in \Bbb{N}^2} A_{n,k}$, which is a countable union of compact subspaces. Since $\Bbb{R}^\omega$ is regular, $\Bbb{R}^\infty$ must be regular. From this we conclude, using the above theorem, that $\Bbb{R}^\infty$ is paracompact.
Does this seem right? I feel like I'm being a bit careless in some spots (e.g., is $A_{n,k}$ compact in $\Bbb{R}^\infty$ or just in $\Bbb{R}^\omega$? Do I need to worry about that?)
I think you're right that the map $\pi^k$ that sends $x \in \mathbb{R}^\infty$ to its first $k$-coordinates in $(x_1, \ldots, x_k) \in \mathbb{R}^k$ (in its standard Euclidean or product topology) is a homeomorphism between $R_k:= \{x \in \mathbb{R}^\omega: \forall n >k : x_n = 0\}$ and $\mathbb{R}^k$. As a restricted projection it will be continuous (under both the product topology and the box opology on $\mathbb{R}^\omega$) and it's also open, as $\pi_k[(\prod_n O_n) \cap \mathbb{R}^\infty] = \prod_{i=1}^k O_i$ (whenever the set we're taking the image of is non-empty (so all but finitely many $O_n$ contain $0$), making $\pi^k$ a continuous open bijection between $R_k$ in the box subspace topology and $\mathbb{R}^k$.
Taking the compact boxes $[-n,n]^k \subseteq \mathbb{R}^k$ and their homeomorphs $A_{n,k} := (\pi^k)^{-1}[[-n,n]^k]$, which thus are also compact in $\mathbb{R}^\infty$ (compactness is absolute) we indeed see that $\mathbb{R}^\infty = \cup_{n,k} A_{n,k}$ (for any $x$ in $\mathbb{R}^\infty$ there is some coordinate $k_x$ beyond which $x_n = 0$, and there is some $N_x$ such that $|x_n| \le N_x$ for all $n< k_x$ which means that $x \in A_{N_x,k_x}$) which is thus $\sigma$-compact and hence Lindelöf $T_3$ and strongly paracompact (also called hypocompact).