About the decidability of sets enumerated in non decreasing order
Suppose there were a computable $f$ as described in the question. Then we could solve the Halting problem as follows.

Given a Turing machine $T$, consider the computable function $g$, defined by
$$g(k) = \begin{cases}
1 &amp; \text{if $T$ halts in $\leq k$ steps} \\
0 &amp; \text{otherwise}
\end{cases}$$
This is a computable, nondecreasing and total function. Therefore, we may use $f$ to establish that the range of $g$ is decidable. But now we just check whether $1$ is in the range to figure out whether $T$ halts.

Notice that I never used the finite/infinite distinction. The above argument works even if we assume that $f$ enumerates a finite set. Are you sure you got your theorem correct? The one I know says that an infinite set is decidable if it can be enumerated by a computable strictly increasing function. There is of course another theorem stating that every finite set can (obviously) be enumerated in increasing order.

Supplemental: Let us look at the matter from a constructive point of view. We have two constructive theorems, assuming Markov principle (which is generally used in computability theory):


  Theorem 1: If a finite set can be enumerated in non-decreasing order then it is decidable.


(A set is said to be finite if it is isomprhic to $\{0, 1, \ldots, n-1\}$ for some $n \in \mathbb{N}$, so the extra conditions in the theorem does not actually help, as every finite set is decidable.)


  Theorem 2: If a non-finite set can be enumerated in non-decreasin order then it is decidable.


However, we cannot just stick these two together into "any set enumerated in increasing order is decidable" because the following is not a constructive theorem:


  Theorem (classical): Every set is finite or non-finite.


In fact, in computability theory there are counter-examples. An immune set is infinite, but it does not contain any infinite computable sequence.