Constructive proof of the Halting Problem
I think if we want to answer this problem constructively, then we should be able propose problem constructively.
Let language of arithmetic be $L=\{0,S,+,\cdot \}$ and $\phi(n,x,y)$ be kleene predicate in $L$. Now the first problem is:


  Is there any $m\in \mathbb{N}$ such that $$H_1(m):=\forall x\in\mathbb{N}:\forall y \in \mathbb{N} \neg \phi(x,x,y)\leftrightarrow \exists z\in \mathbb{N} \phi(m,x,z)$$
  ?


Proof. Suppose there exists such $m$, so by above definition we have:
$$\forall y \in \mathbb{N} \neg \phi(m,m,y)\leftrightarrow \exists z\in \mathbb{N} \phi(m,m,z)$$
and this leads to a contradiction, therefore there does not such $m$. (note that $\neg \exists A(x)\rightarrow \forall x \neg A(x)$ has a constructive proof.)

Define $$H_2(m):=\forall x\in \mathbb{N}\exists ! y\in\{0,1\} \phi(m,x,y) \land \forall x\in\mathbb{N}:\exists u \in \mathbb{N}\phi(x,x,u)\leftrightarrow \phi(m,x,1)$$


  Is there any $m$ such that $H_2(m)$ satisfied in natural numbers?


proof. If there exists such m, we can construct number $m'$ from $m$ such that  $H_1(m')$ satisfied in natural numbers and this lead to a contradiction, therefore we have $\forall n\in \mathbb{N}\neg H_2(n)$.

We can see that $H_2(n)$ means that "partial recursive function with code $n$ is a solution for halting problem". Actually we can not deduce constructively that $$\neg H_2(n)\leftrightarrow \neg(\forall x\in \mathbb{N}\exists ! y\in\{0,1\} \phi(n,x,y)) \lor \neg(\forall x\in\mathbb{N}:\exists u \in \mathbb{N}\phi(x,x,u)\leftrightarrow \phi(n,x,1))$$, but $\neg H_2(n)$ means 
there exists a time in tmeline that we figure out $\neg\forall x\in \mathbb{N}\exists ! y\in\{0,1\} \phi(n,x,y)$ or $\neg\forall x\in\mathbb{N}:\exists u \in \mathbb{N}\phi(x,x,u)\leftrightarrow \phi(n,x,1)$ although maybe we don't know which of them is true this moment.