For the plane $\mathbb{R}^2$ the cross topology is the topology in which a set $\mathcal{O}$ is open iff each point $x\in\mathcal{O}$ is contained in a "cross" formed by a horizontal open interval and vertical open interval within $\mathcal{O}$. More precisely, for $r>0$ and $p=\langle x,y\rangle\in\Bbb R^2$ let $C(p,r)=\Big((x-r,x+r)\times\{y\}\Big)\cup\Big(\{x\}\times(y-r,y+r)\Big)\;;$ $\mathcal{O}$ is open iff for each $p\in\mathcal{O}$ there is an $r_p>0$ such that $C(p,r_p)\subseteq\mathcal{O}$.
Let $A$ be the open first quadrant; clearly the origin is in the closure of $A$, but it's not in the sequential closure of $A$, so the plane with the cross topology is not Fréchet.
To see this, suppose that $\sigma=\langle p_n:n\in\Bbb N\rangle$ is a sequence in $A$, where $p_n=\langle x_n,y_n\rangle$. The cross topology is finer than the Euclidean topology, so $\sigma$ cannot converge to the origin in the cross topology unless it does so in the Euclidean topology. Thus, we may assume that $\langle x_n:n\in\Bbb N\rangle\to 0$ and $\langle y_n:n\in\Bbb N\rangle\to 0$. By passing to a subsequence, if necessary, we may further assume that $\langle x_n:n\in\Bbb N\rangle$ and $\langle y_n:n\in\Bbb N\rangle$ are strictly decreasing. For $n\in\Bbb N$ let $S_n$ be the segment $\overline{p_np_{n+1}}$, and let $S=\bigcup_{n\in\Bbb N}S_n$; $S$ is the graph of a piecewise linear function bijection of $(0,x_0]$ onto $(0,y_0]$ (with countably infinitely many pieces). Let $U=\Bbb R^2\setminus S$; then $U$ is an open neighborhood of the origin in the cross topology that contains no term of $\sigma$, so $\sigma$ does not converge to the origin in the cross topology. $\dashv$
Essentially the same argument shows that a sequence converges to a point $p$ in the cross topology iff it is eventually in every $C(p,r)$. We can use this to show that the plane is sequential in the cross topology.
To see this, suppose that $S\subseteq\Bbb R^2$ is sequentially closed in the cross topology. If $p=\langle x,y\rangle\in\Bbb R^2\setminus S$, no sequence in $S$ converges to $p$, so there is some $r>0$ such that $C(p,r)\cap S=\varnothing$; let $V_0=C(p,r)$. No sequence in $S$ converges to any point of $V_0$, so for each $q\in V_0$ there is an $r_q>0$ such that $C(q,r_q)\cap S=\varnothing$; let $V_1=\bigcup\{C(q,r_q):q\in V_0\}$. In general, if $V_n$ is disjoint from $S$, then for each $q\in V_n$ there is an $r_q>0$ such that $C(q,r_q)\cap S=\varnothing$, and we let $V_{n+1}=\bigcup\{C(q,r_q):q\in V_n\}$. Finally, let $V=\bigcup\{V_n:n\in\omega\}$; the construction ensures that $V$ contains a cross centred at each of its points, so in the cross topology $V$ is an open nbhd of $p$ disjoint from $S$. Thus, $S$ is closed in the cross topology, which is therefore sequential. $\dashv$