If we define a natural number $n$ as $$n\in\bigcap\{x:x\text{ is an inductive set}\}$$ then showing any natural number is either zero or a successor ordinal is equivalent to showing there is an inductive set containing no limit ordinals. How can this be shown?
It's clear that some successor ordinals exist: $\{\emptyset\}$; and that an inductive class with no limit ordinals also exists: $$\emptyset\cup\{\alpha: \alpha \text{ is a successor ordinal}\},$$
yet I still struggle to prove the result.
On
We'll show $$\bigcap\{x:x\text{ is a limit ordinal}\}$$ is an inductive set containing no limit ordinal. Since there is a limit ordinal, this is a set. If $$y\in \bigcap\{x:x\text{ is a limit ordinal}\},$$ then for each limit ordinal $x$, $y\in x$, hence $y\neq x$. This shows the set contains no limit ordinals. Clearly $$0\in \bigcap\{x:x\text{ is a limit ordinal}\}.$$ If $$y\in \bigcap\{x:x\text{ is a limit ordinal}\},$$ then for any limit ordinal $x$, $y\in x$, hence $y+1\in x$, and $$y+1\in \bigcap\{x:x\text{ is a limit ordinal}\}.$$
Assuming you are defining an inductive set as one that contains $\emptyset$ and contains $s(\alpha):=\alpha\cup\{\alpha\}$ whenever it contains $\alpha$, you can reason as follows:
By the axiom of infinity, there is some inductive set $X$. Let $$Y=\{\alpha\in X\mid \alpha \text{ is empty or is a successor ordinal.}\}.$$ Then $\emptyset\in Y$, and if $\alpha\in Y$, then $s(\alpha)$ is a successor ordinal (since $\alpha$ is an ordinal) in $X$ (since $X$ is inductive), and so is also in $Y$, so $Y$ is inductive, and by definition $Y$ consists only of $\emptyset$ and successor ordinals.
Then in conclusion, as you observed in your question, the minimal inductive set $\mathbb N$ is contained in $Y$, hence also consists only of $\emptyset$ and successor ordinals.