Arens–Fort space

Remarks

Every point other than $(0, 0)$ is isolated, and every open set containing $(0, 0)$ meets $X \setminus \set{(0, 0)}$, so $(0, 0)$ is in the closure. Let $(p_k)$ be a sequence in $X \setminus \set{(0, 0)}$ with infinitely many distinct terms, and let $S$ be the set of its terms. Suppose some column contains infinitely many points of $S$, and remove these points from $X$. The rest of $X$ misses infinitely many points of only one column, so it is an open set containing $(0, 0)$. Otherwise every column meets $S$ in a finite set, so $X \setminus S$ is an open set containing $(0, 0)$. In both cases an open set around $(0, 0)$ misses infinitely many terms, so the sequence does not converge to $(0, 0)$. A sequence with only finitely many distinct terms, all different from $(0, 0)$, stays outside the open set $X$ minus these terms. Thus $X \setminus \set{(0, 0)}$ is sequentially closed but not closed. An open set containing $(0, 0)$ is also closed, since its complement misses $(0, 0)$, and each $\set{x}$ with $x \ne (0, 0)$ is open and closed, so the space is regular. A countable regular space is Lindelöf, and a regular Lindelöf space is normal; this gives normality.