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I need a bit more further explanation in part of the proof which states that card(n) = n for all natural numbers n < ω.

Now the proof is done by induction so that for case n+1 one wants to prove that card(n+1) = n+1.

I wonder why is it obvious to claim that card(n+1) ≤ n+1 using the definition

card(x) = min {α| ∃f: α↔ x} is the cardinality of the set x.

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    I need more details in your definition. Does it means card$(x)=\min \{ \alpha : \exists f: \alpha \to x, f \thinspace \text{is a function} \}$? – 2017-01-11
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    @positrón0802 $f$ had better be *surjective* in order for that definition to work. – 2017-01-11
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    Actually f is supposed to be bijection – 2017-01-11
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    @Eurydice In the context of ZFC, both definitions are equivalent, and indeed the one via surjections is easiest to work with in most contexts. – 2017-01-11

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To show that $card(x)\le y$, it's enough to show that there is a bijection from $y$ to $x$ in the first place, since then the least $\alpha$ with a bijection from $\alpha$ to $x$ - that is, $card(x)$ - can't be bigger than $y$.

So: can you think of a bijection from $n+1$ to $n+1$?