Internal Set Theory: Start with nonstandard finite E, then the standard subset of E is infinite?

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I'm working on understanding Edward Nelson's Internal Set Theory.

I wonder if folks can check my reasoning.

I need to reference a result which says that if E is a set then the following two conditions are equivalent.

(1) All the elements of E are standard.

(2) E is standard and finite.

Let $\nu\in\mathbb{N}$ be a nonstandard number and consider the interval $[0,\nu]$ which is finite in the sense that all initial intervals of $\mathbb{N}$ are finite by transfer.

By Standardization, we may form $^S[0,\nu]$ which is standard and it would be tempting to say finite since I wouldn't think it would have any more elements than before. But it cannot be finite because then it would be standard and finite and all its elements would be standard but those would be all the natural numbers you cannot construct the set of only the standard natural numbers within IST. Also, as standard sets $^S[0,\nu]=\mathbb{N}$ and $\mathbb{N}$ isn't finite.

Is this reasoning sound?

EDIT: I tried to focus the question I'm asking.