= Solution
Use the <well-order> of the <natural numbers> to list $A$ in increasing order. Define recursively
$$
f(0)=\min A,\qquad f(n+1)=\min\bigl(A\setminus\{f(0),\ldots,f(n)\}\bigr).
$$
The remaining <set> is nonempty because $A$ is infinite and only a <finite set> has been removed. Each chosen value is larger than the preceding one, so $f$ is an <injective function>.
To prove it is a <surjective function>, fix $a\in A$. If $a$ were never selected, every selected minimum would be at most $a$. This would put infinitely many distinct values into the <finite set> $\{0,\ldots,a\}$, a contradiction. Hence
$$
\boxed{f:\mathbb N\longrightarrow A\text{ is a bijection}.}
$$
The <increasing enumeration of an infinite subset of natural numbers> uses no arbitrary choice: each next element is the uniquely determined minimum.
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