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For a family of functions $f_i:X\to Y_i$ into topological spaces, the initial topology on $X$ is the coarsest topology making them all continuous. Its subbasis consists of $f_i^{-1}(U)$ for open subsets $U\subseteq Y_i$. A map into $X$ is continuous exactly when all its composites with the $f_i$ are continuous.

On the real plane, take the initial topology of all separately continuous real-valued functions. Its coordinate lines have their usual induced topology, and the nonzero rational grid is dense, but its diagonal is uncountable and discrete. Thus the space is separable and nonmetrizable. Rectangular bump functions show that no sequence from the nonzero rational grid converges to the origin, so the space is not first countable either.

Fix a horizontal line $L=\mathbb{R}\times \{c\}$. The restriction of every member of $\mathcal{H}$ to $L$ is continuous in its usual topology. Therefore every subbasic open set intersects $L$ in a usual open set, so the induced topology is no finer than the usual one. Conversely the first coordinate projection is continuous for the initial topology, and its inverse images induce every usual open set on $L$. The two topologies on $L$ are equal. The second projection gives the identical argument for a vertical line.

Write $F={\overline{Y}}^X$. For every nonzero rational $r$, the intersection of $Y$ with $\{r\}\times \mathbb{R}$ is dense in that vertical line, since the nonzero rational numbers are dense in the real line. Hence $\{r\}\times \mathbb{R}\subseteq F$. For each real $c$, it follows that $F\cap (\mathbb{R}\times \{c\})$ contains $(\mathbb{Q}\setminus \{0\})\times \{c\}$ and is closed in that line. This dense subset forces the whole horizontal line into $F$. Thus $F=X$: $Y$ is a countable dense subset. The two successive uses of line density are necessary; the initial topology is finer than the Euclidean topology, so Euclidean density alone would not suffice.

Let $\mathcal{H}$ be the set of separately continuous real-valued functions on the plane. Take as a subbasis all sets $h^{-1}(U)$, where $h\in \mathcal{H}$ and $U$ is open in the usual real line. Unions of finite intersections of these sets form a topology, and every $h$ is continuous for it by construction. Conversely every topology making every $h$ continuous must contain the subbasis, hence this entire generated topology. Thus the required coarsest topology exists and is unique: it is the initial topology of the family $\mathcal{H}$.

The family is a set, being a subset of ${\mathbb{R}}^{{\mathbb{R}}^2}$, so this construction has no proper-class issue. The coordinate projections belong to $\mathcal{H}$; consequently this topology contains the ordinary Euclidean topology. This last observation will also make convergence in the new topology imply ordinary coordinatewise convergence.

Suppose for contradiction that $p_n=(u_n,v_n)\in Y$ converges to the origin in the initial topology. The coordinate projections then give $u_n\to 0$ and $v_n\to 0$ in the usual real line. Rectangular bumps detecting nonconvergence, using anisotropic supports, avoids any unsupported regularity assertion about a radial maximum. SetThe radii are positive since neither coordinate of $p_n$ vanishes. Each rectangular support misses both coordinate axes, so $B$ vanishes on those axes. For a fixed $y=c\ne 0$, all but finitely many vertical support intervals miss $c$, because their centres and radii tend to zero. Hence $B(x,c)$ is a finite maximum of continuous functions of $x$, and is continuous. At $c=0$ it is identically zero. The same argument with the coordinates exchanged proves separate continuity everywhere. Thus $B$ is continuous for the initial topology.

However $B(p_n)=1$ for every $n$, whereas $B(0,0)=0$. This contradicts preservation of convergent sequences by a continuous function. Therefore no sequence from $Y$ converges to the origin.

The radial hint in the scan is not separately continuous for every candidate sequence: for $p_n=(2^{-n},2^{-2n})$, its value at the origin is zero, while its value at $(2^{-n},0)$ is at least $1-2(2^{-2n})/\sqrt{2^{-2n}+2^{-4n}}$, which tends to one. The rectangular construction supplies the needed valid proof independently of that hint.