I want to prove that $\forall n, m \in \mathbb{N}$ $\lvert n \rvert = \lvert m \rvert \iff n = m$ where $\mathbb{N}$ is defined as the smallest inductive set. My proof is like this.
Lemma 1 If $X$ is inductive, $\{\alpha \in X \mid \alpha$ is an ordinal which is not a nonzero limit ordinal$\}$ is inductive.
Proof Let $Y$ be the latter set. It is clear that $0 \in Y$. If $\alpha \in Y$, $\alpha \cup \{\alpha\}$ is a successor ordinal so $\alpha \cup \{\alpha\} \in Y$. Thus $Y$ is inductive. $\blacksquare$
Lemma 2 All nonzero natural numbers are successor ordinals.
Proof By definition of $\mathbb{N}$, $\mathbb{N} \subset \{\alpha \in \mathbb{N} \mid \alpha$ is an ordinal which is not a nonzero limit ordinal$\}$. $\blacksquare$
Since $\alpha + 1 = \beta + 1 \iff \alpha = \beta$, we can define a function $D \colon \mathbb{N} \to \mathbb{N}$ as below.
$D(k) =
\begin{cases}
l & \text{if } k \neq 0\text{, the unique } l \text{ s.t. } l + 1 = k\\
0 & \text{if } k = 0
\end{cases}$
The restriction $D\vert_{\mathbb{N} \setminus 0}\colon \mathbb{N} \setminus 0 \to \mathbb{N}$ is trivially a bijection. Using $D^l$ in the context of iteration, $D^0 = \operatorname{id}_\mathbb{N}$, we obtain the following lemma.
Lemma 3 The least natural number $l$ s.t. $D^l(k) = 0$ is $k$.
Proof $D^0(0) = 0$. Assume that $k \neq 0$ is the least natural number $l$ s.t. $D^l(k) = 0$. Then $D^{k + 1}(k + 1) = D^k(k) = 0$ and $D^k(k + 1) = D^{D(k)}(k) \neq 0$ by the minimality of $k$. $\blacksquare$
Finally we prove a property of $D$ about cardinality.
Lemma 4 $\lvert n \rvert = \lvert m \rvert \Rightarrow \lvert D(n) \rvert = \lvert D(m) \rvert$
Proof Let $f\colon n \to m$ be the bijection. We can construct a new bijection $g\colon D(n) \to D(m)$ by twisting $f$ like below.
$g(k) =
\begin{cases}
f(k) & \text{if } f(k) \neq m \\
f(n) & \text{if } f(k) = m
\end{cases}$
Since $f(k) = f(n) \Rightarrow k = n \not\in D(n)$, $g$ is injective. For all $l \in D(m)$, if $f(n) = l$, $f^{-1}(m) \in D(n)$ so $g(f^{-1}(m)) = f(n) = l$. Otherwise if $f(n) \neq l$, $f^{-1}(l) \in D(n)$ so $g(f^{-1}(l)) = l$. Thus $g$ is surjective, and finally bijective. $\blacksquare$
Now we can prove the theorem.
Theorem $\forall n, m \in \mathbb{N}$ $\lvert n \rvert = \lvert m \rvert \iff n = m$
Proof $\Leftarrow$ side is trivial, so we only have to show the $\Rightarrow$ side. WLOG $n < m$, then $\lvert D^n(n) \rvert = \lvert D^n(m) \rvert$, but $D^n(n) = \emptyset$ and $D^n(m)$ is not empty, leading to a contradiction. $\blacksquare$
Is my proof correct and independent of $\mathsf{AC}$? And is there more simple approach?
At the end of the day, simplicity is all about how you organise your knowledge. And we cannot avoid induction here, at the end of the day it will have to play a role somewhere.
After proving that $\Bbb N$ is the set of all ordinals which are not greater than a limit ordinal (I'm not counting $0$ as a limit ordinal, obviously), which is something pretty straightforward to do, we can start using induction.
Now you can prove that if $m<n$, as ordinals, then there is no injection from $n$ into $m$. This is now enough since we have:
Therefore, if $|n|=|m|$, then it cannot be that $m<n$, or that $n<m$ by a symmetric argument. So it has to be that $n=m$.
Of course, no choice is involved in these arguments whatsoever.