How can I prove that $\operatorname{card}(n)=n$ for all $n \in \omega$, where we defined
$$\operatorname{card}(x):= \min \{\alpha \in \operatorname{Ord} \mid \exists f: \alpha\to x\ \wedge f \text{ is bijection}\}$$
I am trying to use induction and yes for case $n=0$ it is clear. But how to prove the induction step...
Assume the property for all $k\leq n$. Assume there is a bijection $f : n+1 \to k+1$, with $k\leq n$ (obviously, $card(n+1) \neq 0$).
Then either $f(n) = k$, or you can compose $f$ with a bijection $g$ such that $g\circ f(n) = k$. So we can assume without loss of generality that $f(n) = k$. Then it shouldn't be too hard to prove that $f_{\mid n}$ is a bijection $n\to k$, which allows you to use your induction hypothesis, so that $n=k$, and so $n+1= k+1$, which was the desired conclusion.