There are lots of subtleties involved in what constitutes definable (by finite amount of information). For example, one might be tempted to say the following: an element $a$ of some structure $\mathcal{M}$ is definable iff there exists a (first-order, parameter-free) formula $\varphi(x)$ s.t. $\mathcal{M} \models \varphi(x)$ iff $x = a$. But under this definition it actually could be the case that all real numbers (in fact, every mathematical object whatsoever) are definable. See, for example, https://arxiv.org/abs/1105.4597 . This does not contradict the countability arguments because the proof that the set of real numbers is uncountable is carried out within the model of ZFC, while the argument that the set of definable real numbers is countable is carried out within the metatheory. So, while the set of real numbers within the said model $\mathcal{M}$ (of ZFC) can be put in bijection with the set of natural numbers, this bijection is simply not contained in $\mathcal{M}$ itself, so it does not contradict the theorem that the set of real numbers are uncountable, proved in $\mathcal{M}$.
The idea behind this is actually not that complicated. Assuming $\mathrm{V} = \mathrm{HOD}$, there exists a definable well-ordering of the universe. So this allows us to pick, in a definable way (by choosing the least element in a fixed definable well-ordering of the universe), an element $x$ that satisfies $\exists x \varphi(x)$ for any such formula that is true in $\mathrm{V}$. Running through the proof of downward Lowenheim-Skolem using this fact then gives the required model. (The precise details are in the paper I cited above.) Assuming the existence of a definable well-ordering of the universe and that every natural number in the universe is definable (which is probably not trivial - the universe could contain plenty of non-standard natural numbers) also completes the argument of @Vincent in the comments. If one can prove that a (definable - otherwise the remainder doesn’t make much sense) set $A$ is countable in $\mathcal{M}$, i.e., there is a bijection between $\mathbb{N}$ and $A$ within $\mathcal{M}$, then we can define any element of $A$ by choosing the least bijection $A \to \mathbb{N}$ according to the well-ordering of the universe, and then use the induced definable numbering of $A$ and the fact that all elements of $\mathbb{N}$ are assumed definable to define elements of $A$, showing that any set that’s countable in-universe must have all its elements definable, in this scenario.
But then again, as the example with all objects being definable show, you cannot actually go back - all elements of a set are definable only imply it is countable looking from outside, not necessarily in-universe. To summarize, you have the following:
- If you can prove (necessarily outside the universe) that all elements of a set $A$ are definable, then $A$ must be countable looking from outside the universe. But it could be uncountable in-universe;
- Under some assumptions (the existence of a definable well-ordering of the universe and all natural numbers are definable), you can prove (again necessarily outside the universe), if (a definable) $A$ is countable within universe, then all elements of $A$ are definable. However, you can’t prove this if you only know $A$ is countable from outside the universe. (This follows from downward Lowenheim-Skolem and the assumption that there exists a universe in which some elements are not definable.)