How to prove that $\mathbb R^\omega$ with the box topology is completely regular

Seeing that there is no elaborated answer, I've decided to give a go at it.

Lemma: If $h:\mathbb{R}^\omega\to\mathbb{R}^\omega$ (box topologies) is defined by $h(x_1,x_2,...):=(a_1x_1+b_1,a_2x_2+b_2,...)$ where $a_i>0$, then $h$ is a homeomorphism. The proof is here.

Main proof: Let $x\in \mathbb{R}^\omega$ and $A$ is closed subset that does not contain $x$. By lemma, we may WLOG assume that $x$ is origin. Moreover, we can find a basis element of the form $(-c_1,c_1)\times(-c_2,c_2)\times...$ that does not intersect the closed subset $A$. By lemma again, we may WLOG assume that $c_i=1$ for all $i$.

Now, $\mathbb{R}^\omega$ instead under the uniform topology is metrizable and hence completely regular (in fact normal). Given the origin and the closed set $C:=\mathbb{R}^\omega - ball_0(1)$, we can find a continuous $f:\mathbb{R}^\omega\to[0,1]$ such that $f(0)=0$ and $f(C)=1$. Since the box topology is finer than uniform topology, the same $f$ is continuous if $\mathbb{R}^\omega$ is instead given the box topology.

Lastly, since $ball_0(1)\subseteq(-1,1)^\omega$, we have $A\subseteq C$ so $f$ is our desired continuous function to prove that $\mathbb{R}^\omega$ under box topology is completely regular.

Remark: The lemma and this proof can be easily generalized to the case where we have an arbitrary index set $J$ instead of $\omega$.

First, it suffices to only consider the case where $a = (0, 0, \dots)$ and the open neighborhood $(-1,1)^\mathbb{N}$ of $a$ is disjoint from $F$ (why?).

Hint: Now, having reduced the general case to this one, note that the uniform topology on $\mathbb R^\mathbb{N}$ is coarser than the box topology. Hence any function continuous on $\mathbb R^\mathbb{N}$ in the uniform topology is also continuous with respect to the box topology.

What would be a canonical choice for your function?