$\mathbb {R}^{\omega}$ in the box topology is not first countable.

Use a diagonal argument. Given a putative countable basis at x (wlog x = all zeros), construct a basic open set not containing any of the countable ones by making sure it's smaller than the $i^{\mathrm{th}}$ open set in the $i^{\mathrm{th}}$ coordinate.

If it is 1st countable then for any subset A with a limit point x then there should exist a sequence $x_n$ converging to x. Now take $A=\{(y_n)_n: y_n>0 \forall n\}$. 0 is a limit point of A but there is no sequence in A converging to 0. If possible let $a_n$ is any such sequence where $a_n=(a_{n1},a_{n2},\ldots)$ then the open set $(-a_{11},a_{11})\times (-a_{22}, a_{22})\times\cdots$ contains no points in the sequence.