Arens space $S_2$ is sequential but not Fréchet–Urysohn

Every neighbourhood of $\infty$ meets $A$, so $\infty\in\overline A$ by [L1]. No sequence in $A$ converges to $\infty$: if it visits a row infinitely often, a neighbourhood omitting that row defeats convergence. If it visits every row finitely, use [L2] to put the threshold on each visited row one above the maximum selected second coordinate, and threshold $0$ on every unvisited row. The resulting neighbourhood omits the whole sequence.

Let $C$ be sequentially closed. An isolated closure point lies in $C$. If $x_n\in\overline C$, then $C$ meets its $n$-th row arbitrarily far out; recursion and least elements from [L2] give a sequence of row points in $C$ converging to $x_n$, so $x_n\in C$. If $\infty\in\overline C$, then infinitely many $x_n$ lie in $C$: otherwise omit the finitely many rows whose centres lie in $C$. In every remaining row, the preceding conclusion shows that $C$ has only finitely many points; using their maximum as in [L2] gives a canonical tail disjoint from $C$. These tails form a neighbourhood of $\infty$ disjoint from $C$, contradicting $\infty\in\overline C$.

Hence $S_2$ is not Fréchet–Urysohn.

The indices $n$ with $x_n\in C$ form an infinite subset of $\mathbb N$; list them increasingly using [L2]. The resulting sequence of row centres converges to $\infty$, so sequential closedness puts $\infty$ in $C$. Thus every sequentially closed $C$ contains all its closure points and is closed. Therefore $S_2$ is sequential.

The two conclusions prove the example.