Okay, so a couple of things.
Suppose that you know the index $e$ such that $\varphi_e$ is the indicator function for a nonempty recursive set $A$ (so we are promised that $\varphi_e$ is total recursive and not identically zero). Then:
- If you don't mind repeating elements, take $f(e,n) = n$ if $\varphi_e(n) = 1$ and $f(e, n) = f(e, n-1)$ otherwise, while $f(e, 0) = \mu x(\varphi_e(x) =1)$.
- If you don't like repeating elements in your enumeration of infinite recursive sets, you are out of luck. Determining whether $\varphi_e$ is eventually $0$ (i.e., $A$ is finite) is not a computable task, nor is determining whether it is identically zero. You could solve the halting problem if you could solve this.
This function of course doesn't converge for most $n$ when $\varphi_e$ is not total recursive (or when $\phi_e$ is never $1$). If you need $f(e,n)$ to be an increasing enumeration when $\varphi_e$ is total recursive and the indicator function for a nonempty recursive set, and also be a total recursive function, you are again out of luck. Once again this would solve the halting problem.
We have to distinguish between sets and definitions of sets.
A set is decidable (or computable, or recursive - the subject has unfortunately redundant terminology!) iff some Turing machine decides it. Every finite set is decidable since we can always "hard-code" a Turing machine to accept a given finite set: fixing $a_1,...,a_n$, just write a program which on input $k$ checks whether $k=a_1$, whether $k=a_2$,... , whether $k=a_n$, and outputs "YES" if the answer to one of these questions is YES and outputs "NO" otherwise.
However, some definitions of very simple sets appear computationally intractable. For example, for any sentence $\varphi$ let $$True_\varphi=\{x: (x=0\mbox{ and $\varphi$ is true})\mbox{ or } (x=1\mbox{ and $\varphi$ is false})\}.$$ Obviously $True_\varphi$ defines a decidable set (either $\{0\}$ or $\{1\}$), but we don't know which. This suggests the following notion (the terminology below is my own, I don't know if there's a more common one):
Say that a definition $\delta$ of a set of natural numbers is concrete iff there is some Turing machine $M$ such that $\mathsf{ZFC}$ (say) proves that $M$ decides the set defined by $\delta$.
(I'm being a bit vague here about what exactly I mean by "definition" - it doesn't really matter, but if you like we can take it to mean "formula in the language of set theory with one free variable which $\mathsf{ZFC}$ proves only ever holds of natural numbers.")
Before moving on, let me give a quick caveat:
Not every decidable set has a concrete description!
The concretely-describable sets form a proper subclass of the decidable sets; every finite set is indeed concretely describable, though, so for now this is a reasonable thing to think about. (See the end of this answer for more on this.)
The point, then, is the following:
Not every definition of a finite set need be concrete.
In particular, "The set containing exactly the $7918$th value of the Busy Beaver function" is not a concrete definition of a set; however, the set it defines is finite and does in fact have some concrete definition, we just don't know what it is.
OK, now let me get back to the point above about concrete describability vs. decidability.
Since we can search through $\mathsf{ZFC}$-proofs (and assuming $\mathsf{ZFC}$ is sound), we can compute an enmeration $(M_i)_{i\in\mathbb{N}}$ of all Turing machines which $\mathsf{ZFC}$ proves decide a set - that is, which $\mathsf{ZFC}$ proves always halt on all inputs. But now we can diagonalize against these: the set $$D_{\mathsf{ZFC}}=\{i: M_i(i)=0\}$$ is decidable but has no concrete description.
It's a good exercise to check why the above definition of $D_{\mathsf{ZFC}}$ is not in fact concrete! Also, note that $\mathsf{ZFC}$ isn't really important here - we can replace $\mathsf{ZFC}$ throughout with any other "appropriate" theory, like first-order Peano arithmetic.
In particular, "The set of numbers decided by machine $M$" is not concrete in general ... since maybe $\mathsf{ZFC}$ doesn't prove that $M$ actually decides a set!
This is one of many situations in computability theory where semidecidability (or computable enumerability, or recursive enumerability) is a much tamer notion than decidability: while it's hard to tell whether a Turing machine decides a set, every Turing machine definitely accepts a set, so "is $\mathsf{ZFC}$-provably equivalent to a definition of the form 'the set of numbers accepted by $M$'" is a much better-behaved notion of concreteness than "is $\mathsf{ZFC}$-provably equivalent to a definition of the form 'the set of numbers decided by $M$.'"
(Incidentally, this winds up connecting with the topic of what functions a given theory - such as $\mathsf{ZFC}$ - can prove to be total. There's a vast literature on this, mostly in proof theory but spilling into computability theory and the study of models of arithmetic, but for a starting point see Beklemishev, Induction rules, reflection principles, and provably recursive functions.)
Best Answer
Consider the following set: $X=\{0\}$ if the Riemann hypothesis is true, and $X=\{1\}$ otherwise. Then:
I can write down a pair of Turing machines $T_0$ and $T_1$, and prove that either $T_0$ computes $X$ or $T_1$ computes $X$.
Of course, I can't tell which machine computes $X$ . . .
. . . but I don't care: there is one, so $X$ is recursive.
If you understand this example, the case of arbitrary finite sets is no different: we can find a computable sequence of Turing machines $T_{e_i}$ ($i\in\omega$) such that $T_{e_i}$ computes $F_i$, where $\{F_i: i\in\omega\}$ is some enumeration of all finite sets. Then if we know $X$ is finite, we know some $T_{e_i}$ computes $X$. We don't know which . . . but we don't care.
Actually, "we don't care" is putting it a bit strongly. We absolutely do care when we're not looking at just one finite set in a vacuum. For instance, let $K_n=K\cap\{0, . . . , n\}$. Then each $K_n$ is computable, but the $K_n$s are not uniformly computable - there is no computable $f$ such that $T_{f(n)}$ is a Turing machine computing $K_n$. Uniformity issues are indeed a Big Deal. But, if you're just asking whether an individual finite set is computable or not, they don't come up.