Box topology first countability

Question

I want to show that $\mathbb{R}^2$ with the topology $\tau_\mathcal{B}$ generated by the base $\mathcal{B} = \{U \times V | U \in \tau_1, V \in \tau_2\}$ where $\tau_1$ and $\tau_2$ are topologies for $\mathbb{R}$ with $\tau_1$ not first countable, is not first countable without using sequences.

Answer 1

For convenience let $X$ denote $\Bbb R$ with the topology $\tau_1$, and let $Y$ denote $\Bbb R$ with the topology $\tau_2$. Let $x\in X$ have no countable local base, and let $y\in Y$ be arbitrary. Let $p=\langle x,y\rangle\in X\times Y$, and suppose that $\{B_n:n\in\Bbb N\}$ is a countable local base at $p$. For each $n\in\Bbb N$ there are $V_n\in\tau_1$ and $W_n\in\tau_2$ such that $p\in V_n\times W_n\subseteq B_n$, so $\mathscr{B}_p=\{V_n\times W_n:n\in\Bbb N\}$ is also a countable local base at $p$.

$\{V_n:n\in\Bbb N\}$ is a countable family of open nbhds of $x$ in $X$, so it is not a local base at $x$, and there is an open nbhd $V$ of $x$ such that $V_n\setminus V\ne\varnothing$ for each $n\in\Bbb N$. But then $V\times Y$ is an open nbhd of $p$ that contains no member of $\mathscr{B}_p$, which therefore cannot be a local base at $p$. This contradiction shows that $X\times Y$ cannot be first countable at $p$.