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I think it is just the usual order on the rational numbers, but since he uses only some of the rationals, it does amount to using ordinals.

But, I think your solution is off by at least a factor of omega.

I see, he has limit ordinals in his embedded order with the sequence 1 - (1/2)^n, with 1 being the equivalent of omega.

You probably even could get an embedded omega squared with

1 - (1/2)^n - (1/3)^m with m>n

So 1/2 = omega

3/4 = 2 * omega

7/8 = 3 * omega

and 1 is omega * omega

He might have omega squared in his version, but I didn't read it that carefully.

Now play the version where the secret ordinal is an uncountable ordinal or a strongly inaccessible cardinal.