TITLE: Non-constructive proofs of decidability?
QUESTION [17 upvotes]: Are there examples of sets of natural numbers that are proven to be decidable but by non-constructive proofs only?

REPLY [3 votes]: Indeed, there exists some mathematical theorems with non-constructive proofs which cannot have any constructive proof (now and ever). Here is an example:
Definition. Let $\boldsymbol\varphi_1,\boldsymbol\varphi_2,\boldsymbol\varphi_3,\cdots$ be a list of all unary partial computable functions (from $\mathbb{N}$ to $\mathbb{N}$). Denote the Chaitin-Kolmogorov Complexity by $\mathcal{C}$ which is defined by $\mathcal{C}(n)=\min \{e\mid\boldsymbol\varphi_e(0)\!\downarrow = n\}$.
Let us note that $\mathcal{C}$ is a total function ($\mathcal{C}:\mathbb{N}\rightarrow\mathbb{N}$).
The following theorem has a non-constructive proof:
Theorem 1. For every natural $n$ there exists some $m$ such that $\mathcal{C}(m)\!>\!n$. That is $\forall y \exists x \mathcal{C}(x)\!>\!y$.
Proof: For each $i\leqslant n$ put $\alpha_i=\boldsymbol\varphi_i(0)$ if the value exists, i.e., $\boldsymbol\varphi_i(0)\!\downarrow$; otherwise let $\alpha_i=0$. Take $m$ to be any natural number greater than all $\alpha_i$'s.
 $\texttt{QED}$
As a matter of fact, the above theorem cannot have any constructive proof:
Theorem 2. There exists no computable function $f$ such that for every natural $n$ one has $\mathcal{C}\big(f(n)\big)\!>\!n$.
Proof: If there were such a computable (and total) $f$ then by Kleene's Recursion (aka Fixedpoint) Theorem there would exist some $\mathbf{e}$ such that $\boldsymbol\varphi_{\mathbf{e}}(0)=f(\mathbf{e})$. For this $\mathbf{e}$ then we would have $\mathcal{C}\big(f(\mathbf{e})\big)\leqslant\mathbf{e}$ contradicting the assumption of $\forall n: \mathcal{C}\big(f(n)\big)\!>\!n$.  $\texttt{QED}$
Now, we have an example of a decidable set whose decidability cannot be proven constructively:
Let $\mathfrak{B}=\{n\in\mathbb{N}\mid \exists m: \mathcal{C}(m)>n\}$. We know that $\mathfrak{B}$ is decidable just because $\mathfrak{B}=\mathbb{N}$. But if anyone (now or in future) can have any constructive proof of its decidability, s/he should be able to compute the characteristic function $\chi_{\mathfrak{B}}$ of $\mathfrak{B}$ (defined by $\chi_{\mathfrak{B}}(x)=0$ if $x\not\in\mathfrak{B}$ and $\chi_{\mathfrak{B}}(x)=1$ if $x\in\mathfrak{B}$). That would amount to showing that $\chi_{\mathfrak{B}}=\mathbf{1}$ the constant function $1$. Thus any constructive proof of the decidability of $\mathfrak{B}$ would show (constructively) that $\mathfrak{B}=\mathbb{N}$ and thus would provide a computable function $f$ such that $\forall n\!\in\!\mathbb{N}: \mathcal{C}\big(f(n)\big)\!>\!n$; contradicting Theorem 2. 
So, I believe, $\mathfrak{B}$ is a decidable set such that no proof of its decidability can be constructive.