 Solution

Let $\kappa =|\alpha |$ be the infinite initial ordinal of the given ordinal. A successor adds only one point, so $|\alpha +1|=\kappa$: one can absorb the additional point by shifting the initial copy of $\omega$ and fixing the rest.

The successor construction from part (a) partitions the new range into the old range, a new row of size $|\alpha +1|$, and a new column of size $|\alpha |$. By hypothesis each part has cardinality at most $\kappa$, and the old range already has cardinality $\kappa$.

A finite disjoint union of copies of a well-orderable infinite cardinal has that same cardinality without invoking the square theorem being proved. Explicitly, write each $\xi <\kappa$ uniquely as $\omega \eta +n$ with $n<\omega$. Interleave three copies byThis stays below $\kappa$ because an infinite initial ordinal is a limit ordinal and is closed under finite successor steps below it. It is injective by uniqueness of the quotient and finite remainder. The reverse injection is immediate, so the Cantor-Schröder-Bernstein theorem proves the finite-union claim.

Applying that bound to the three range pieces yieldsOnly the well-orderability of ordinals, not a choice principle for arbitrary families of sets, was used.