= Solution
Use three nested <sequences>:
$$
A_3=\left\{\left(\frac1n+\frac1{n^2m},\frac1{n^2m^2k}\right):n,m,k\geq1\right\}.
$$
For fixed $n,m$, letting $k\to\infty$ produces the point $(1/n+1/(n^2m),0)$. Letting $m\to\infty$ produces $(1/n,0)$, and letting $n\to\infty$ produces $(0,0)$. Accordingly, put
$$
B=\left\{\left(\frac1n+\frac1{n^2m},0\right):n,m\geq1\right\},\qquad
C=\{(1/n,0):n\geq1\}.
$$
The successive derived sets are
$$
\boxed{A_3'=B\cup C\cup\{(0,0)\},\quad A_3''=C\cup\{(0,0)\},\quad A_3^{\prime\prime\prime}=\{(0,0)\}.}
$$
To see that no limits are missing, take any <convergent sequence> of distinct points from $A_3$. If $n$ is unbounded, a subsequence has $n\to\infty$ and converges to the origin. Otherwise fix $n$ on a subsequence. Unbounded $m$ then forces limit $(1/n,0)$; with $n,m$ fixed, distinctness forces unbounded $k$, giving the corresponding point of $B$. The same argument, with one index fewer at each stage, proves the next two equalities. Distinct $n$-clusters are separated away from the origin because their first coordinates lie in the disjoint <intervals> $[1/n,1/n+1/n^2]$.
The original set has positive second coordinates, the first derived set has isolated $B$-points on the axis, the second has infinitely many $C$-points, and the third is a singleton. All three required successive inequalities follow. This is a <three-stage accumulation tower>.
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