= Solution
Choose an <injection> $j:A\to\mathbb N$. For each $b\in B$, its fiber under the <surjection> $f$ is nonempty, so define
$$
h(b)=\min\{j(a):f(a)=b\}.
$$
Two distinct fibers cannot contain the same label $j(a)$, because $j$ is an <injection> and $f(a)$ has a single value. Thus $h:B\to\mathbb N$ is an <injection>, proving that $B$ is a <countable set>. Taking the least label avoids any need for a separate choice of one element from each fiber. If $A$ is empty, surjectivity makes $B$ empty as well.
For the unheaded enumeration request, list pairs of positive <natural numbers> by their sum. An explicit <Cantor pairing function> is
$$
\boxed{\pi(m,n)=\frac{(m+n-2)(m+n-1)}2+n.}
$$
For $d=m+n-2\ge0$, the positive <integers> $n=1,\ldots,d+1$ give the consecutive labels $d(d+1)/2+1,\ldots,(d+1)(d+2)/2$. These blocks are disjoint, consecutive and cover $\mathbb N$. Consequently $\pi$ is a <bijection> from $\mathbb N\times\mathbb N$ to $\mathbb N$, so \b[$\mathbb N\times\mathbb N$ is countable].
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