Let $(\mathbb{R}^\omega,\tau)$ be the box topology space. I have found that everyone proofs that this topology is not first countable by finding a contradiction, but I thought of a constructive way. Would appreciate comments if I'm missing something. Goes like this:

Let $x\in\mathbb{R}^\omega$ and let $B(x)$ be a local basis for $x$. Then take $a\_{i\varepsilon}=x\_i-\varepsilon$ and $b\_{i\varepsilon}=x\_i+\varepsilon$, and let
$$
I\_\varepsilon=\prod\_{i\in\mathbb{N}}\,
 (a\_{i\varepsilon},b\_{i\varepsilon})
$$
So, in particular, we get that $\forall\varepsilon\in (0,1),\,\,$ $I\_\varepsilon$ are neighbourhoods of $x$. So $\forall\varepsilon\in (0,1)$ $\exists B\_\varepsilon\in B(x)$ s.t. $x\in B\_\varepsilon\subseteq I\_\varepsilon$. Thus there is at least a continnum of elements in $B(x)$, i.e., $|B(x)|\ge c$ and thereby $\mathbb{R}^\omega$ is not first countable.