Natural numbers in set theory is $\{0,1,2,...\}?$

Instead of working "backwards", work "forward". In that case, the need for the axiom of infinity is also mitigated.

Begin by defining ordinals, or cardinals whichever you like. Ordinals work better since they have an inherent successor operation. Now define the notion of "finiteness", it is true that many times this notion makes an appeal to the natural numbers but there are purely set theoretical definitions of finiteness. For example Dedekind-finiteness, Tarski-finiteness, and others.

Now define $\omega$ as the collection of all finite ordinals. This is exactly what you get when you begin with $\{0\}$ and close it under the successor operation. Next show, if you like, that this set is inductive and that it is in fact minimal.

In the course I'm TA'ing right now, the professor (who is a distinguished set theorist) is taking this sort of approach. The natural numbers are some atomic object at first, and after defining the ordinals we take the finite ordinals, $\omega$ as the set theoretic natural numbers. This circumvents the need to talk about inductive sets at all.


Question (quote) "I want to show that $\mathbb{N}$ does not contain an element apart from 0 that is not the successor of anything else in $\mathbb{N}$."

Short Answer This would contradict the minimality of $\mathbb{N}$. You mentioned a correct proof sketch by contradiction.

Long Answer You asked if it can be proved directly; we will show that every element of $\mathbb{N}$ is either $0$ or an iterated-successor of $0$ (i.e. $0$ is an element). This answers your overall question of why the alternative notation $\{0,1,2,\ldots\}$ is justified.

Remark "$\ldots$" is nothing but a symbol---it is meaningless and only used in notations. It's much like "$\infty$" in that regard. Another note is that the ZFC axioms do not restrict us on how to denote our sets. As long as we're clear about which unique set we refer to, we can use any notation for any set; i.e. $\omega$ and $\aleph_0$ are notations for the exact same set even though they're used in different contexts.

Recall

$\mathbb{N}$ is the minimal inductive set; i.e. it's an inductive set such that for each inductive set $B$, if $B\subseteq \mathbb{N}$, then $B=\mathbb{N}$.

Brief review

$A$ is an inductive set if $\varnothing\in A$ and for each $X\in A$, $X\cup\{X\}\in A$.

A natural number is an element of $\mathbb{N}$.

We will use "$+1$" to denote successor; i.e. $X+1=X\cup\{X\}$. From this we can derive the following (induction)

For each formula $\phi(x)$, if $\phi(0)$ holds and for each natural number $k$, if $\phi(k)$ holds, then $\phi(k+1)$, then for each natural number $n$, $\phi(n)$ holds.

At this point most would be satisfied that the notation $\{0,1,2,\ldots\}$ is justified, but we want to be pedantic and use the above theorem

Define the formula $\phi(x)$ by$$\phi(x)\overset{\mathrm{def}}{=} (x\ne 0)\Rightarrow (0\in x)$$

This is something that can be proved by induction for each natural number.

(sorry for not posting proofs, but you already have one)