Let $G = \{0,1,2,\ldots | \}$. Then, obviously, also $G = \{1,2,\ldots| \}$. Because 0 is dominated by the other options.
Now let $S$ be the set of options I can 'cross out'. Clearly, $0\in S$ and if all numbers from $0$ to $n-1$ belong to $S$, then $n$ belongs to S (because $n+1$ is also an option). Thus by strong induction S = N and left has no moves.
Where have I made a mistake?
This is a good question, and if I remember correctly, On Numbers and Games actually says something incorrect/misleading about this idea.
It's true that the set of dominated options of $\omega=\{0,1,2,\ldots\mid\}$ is $\mathbb N=\{0,1,2,\ldots\}$. However, the usual theorem is that you can delete a single dominated option. By a little induction argument, you can delete finitely many of them, but that's all you can do in general.
To quote my abandoned blog on Combinatorial Game Theory, it's not even true that we can delete only finitely many options of $\omega$. As long as we leave an infinite bunch of options, it'd still be equal to $\omega$. For example, we could delete all of the non-squares, and find that $\omega=\{0,1,4,9,\ldots\mid\}$.