Yes we do.

The problem of a chain isn't intended to be limited to the problem of exactly one chain, and I didn't want to complicate the diagram or confuse my readers by showing them a copy of the rationals with each rational replaced by a copy of the integers. If you can't get rid of a larger structure that has a chain in it, you can't get rid of the chain. To put it another way, showing that the chain depicted implies further extra elements isn't the same as ruling out the existence of that chain.

Hence the wording, "How do we get rid of the chain?" not "How do we get rid of this particular exact model here?"

A very quick way to see that there must be more than one chain is to note that if x > y, then x + z > y + z. An element of the nonstandard chain is greater than any natural number, so if we add two nonstandard numbers together, the result must be greater than the nonstandard starting point plus any natural number. Therefore there must be another chain which comes after the first one. For more on this see the linked paper.

EDIT: Several others reported misinterpreting what I had in the original, so I've edited the post accordingly. Thanks for raising the issue, Ilya!