There are two related posts but I didn't see the satisfied answer there:
My question is how to prove infinite set has cardinality greater or equal to $\mathbb{N}$ without axiom of choice, and for generality without using contradiction?
Since $\mathbb{N}$ was constructed through the inductive sets, I can hardly see a direct construction of a function $f$.
The possible solution I have right now seemed to be using the recursion theorem and define $g(f_n,n)$ to be the elements of $B-\{f_0,...,f_n\}$, but this seem to be impossible without AC.
Another way I thought about is argue that one can pick a sequence $b_0,...,$, such that $b_i\neq b_j$ for $i\neq j$. The sequence won't end other wise $B$ will be finite. But this required the usage of contradiction.
How to ptove the theorem without axiom of choice and the usage of contradiction?