I am going to answer twice. Once as if your use of the word "constructive" meant "computable", and once as if it actually meant "constructive" in the sense of "constructive mathematics".

**First answer:** Suppose there were a computable method $c$ of transforming (codes of) computable enumerations of non-empty sets to non-decreasing computable enumerations of non-empty sets. Then we can solve the Halting oracle as follows. Given any Turing machine $T$, consider the sequence $e : \mathbb{N} \to \mathbb{N}$ defined by
$$e(n) = \begin{cases}
1 & \text{$T$ does not halt at step $n$} \\\\ 
0 & \text{$T$ halts at step $n$}
\end{cases}$$
The map $e$ enumerates the set $\{0,1\}$ or $\{1\}$, depending on whether $T$ halts.
By assumption $c$ transforms $e$ into a non-decreasing enumeration $e'$ which enumerates the same set as $e$. If $e'(0) = 0$ then $T$ halts, otherwise it does not.

**Second answer:** In constructive mathematics we can just drop the adjetive "computable" from "computable enumeration". Suppose for every enumeration $e : \mathbb{N} \to \mathbb{N}$ there existed another enumeration which listed the same elements in non-decreasing order. We can derive from this the non-constructive principle [LPO][1] as follows. Consider a map $f : \mathbb{N} \to \{0,1\}$. We are supposed to decide whether $\exists n \in \mathbb{N} . f(n) = 0$. By assumption there is a non-decreasing map $e' : \mathbb{N} \to \mathbb{N}$ which has the same image as $f$. We simply look at $e'(0)$ to tell whether $f$ attains $0$.


  [1]: http://ncatlab.org/nlab/show/principle+of+omniscience