$\Bbb R^{\omega}$ in the box topology is not metrizable

The structure of the proof is as follows: So we have the set $A$ of the all positive sequences.

Then we observe that $0 \in \overline{A}$, as every box-open neighbourhood of $0$ intersects $A$.

So if $\mathbb{R}^\omega$ were metrisable, there is a sequence $x_n$, where all $x_n \in A$ such that $x_n \to x$. But then we show that no sequence from $A$ can converge to $0$ at all, contradiction.

The no-convergence part is a classical diagonalisation argument (literally, here, as well):

Suppose $x_n \in A$ for all $n$, so we write (as each point in $A$ is a sequence of (strictly positive) reals:

$$x_n = (x^{(n)}_1,x^{(n)}_2, x^{(n)}_3, \ldots)$$

where I choose a slightly different notation: the upper $n$ indicates it's the $n$-th point in the sequence from $A$, the lower index is just the coordinate index. As said, being in $A$ means $\forall n,m: x^{(n)}_m > 0$ so we can define the following box-open set:

$$O = \prod_{i=1}^{\infty} ( -x^{(i)}_i, x^{(i)}_i)$$

which contains the point $0 \in \mathbb{R}^\omega$, as each coordinate contains $0$ being a symmetric neighbourhood $(-a,a)$ around $0$.

Then for each $n$, $x_n \notin O$: at the $n$-th coordinate we have $x^{(n)}_n \notin (-x^{(n)}_n, x^{(n)}_n)$, so $x_n \notin O$.

So $0$ has a neighbourhood that contains no point of the sequence, while if $_n \to 0$, by definition of convergence, every neighbourhood of $0$ contains all but finitely many points of the sequence. So indeed $x_n \not\to 0$ in the box topology. This final contradiction then shows that the box product fails the sequence lemma for a specific $A$ and $0$, and this is enough to refute the metrisability (or even first countability) of the box product.

BTW, $0$ is not a special point, the space is not first countable at any point. Taking this $A$ and $0$ makes the proof easier to follow, I think.

I hope this clarifies your questions?