= Solution
Suppose for contradiction that $p_n=(u_n,v_n)\in Y$ converges to the origin in the <initial topology>. The <projection map>[coordinate projections] then give $u_n\to0$ and $v_n\to0$ in the usual <real line>. <Rectangular bumps detecting nonconvergence>, using anisotropic supports, avoids any unsupported regularity assertion about a radial maximum. <Set>
$$
\phi(t)=\max(1-|t|,0),\qquad r_n=|u_n|/4,\quad s_n=|v_n|/4,
$$
$$
b_n(x,y)=\phi\!\left(\frac{x-u_n}{r_n}\right)\phi\!\left(\frac{y-v_n}{s_n}\right),\qquad B(x,y)=\sup_n b_n(x,y).
$$
The 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 $y=c\ne0$, all but finitely many vertical support intervals miss $c$, because their centres and radii tend to zero. Hence $B(x,c)$ is a <maximum>[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 \b[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.
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