It is provable from $\sf ZF$ that there is a surjection from $\Bbb R$ onto $\{A\subseteq\Bbb R\mid A\text{ is countable}\}$. This is because it is provable that the cardinality of $\Bbb R$ is equal to the cardinality of $\Bbb{R^N}$, and then we can just map each sequence of real numbers to the set of numbers appearing in the sequence.
And of course, there is a surjection from $\{A\subseteq\Bbb R\mid A\text{ is countable}\}$ onto $\Bbb R$, simply map $\Bbb N\cup\{r\}$ to $r$ for all $r\notin\Bbb N$; and $\Bbb N\setminus\{n\}$ to $n$ for all $n\in\Bbb N$, and any other countable set to $0$.
But it is provable that if there is a bijection between $\Bbb R$ and $\{A\subseteq\Bbb R\mid A\text{ is countable}\}$, then there is a set which is not Lebesgue measurable. Since it is consistent with $\sf ZF$ that all sets of reals are Lebesgue measurable, it follows that it is consistent that there is no bijection between these two sets.
Note, by the way, that the second surjection onto the reals has an easy injective inverse. If we could find an injection from $\{A\subseteq\Bbb R\mid A\text{ is countable}\}$ to $\Bbb R$, then the Cantor–Bernstein theorem (which requires no choice for its proof) would imply the existence of a bijection. So the problem is indeed finding an inverse to the first surjection.
In other words, it is impossible to uniformly assign every countable set of reals an enumeration without using the axiom of choice.
More generally, however, note that since the axiom of choice implies that given two sets with mutual surjections there is a bijection between them, it is generally not a trivial task to imagine "how such two sets which contradict this" would look like. We are just lucky enough that in this case we can find these sets in the realm of the real numbers. But generally, counterexamples to the axiom of choice are as intangible as its consequences (e.g. a well-ordering of the real numbers).
