As the title states, how do you prove that $A$ is productive? With $W_{x}$ I mean the set of points in which the turing machine with index $x$ halts.
The standard approach that I follow is functional reduction: first I construct a partial recursive function $\psi(x,y)$. Then through the s-m-n theorem I build a partial recursive function $\varphi_{g(x)}(y) = \psi(x,y)$ with $g$ total recursive. Then I proceed to show that $\overline{K}$ reduces to the set I'm studying through $g$. The problem is that in this set the condition includes the index of the partial function ($x$) so I don't know how to construct a suitable $\psi$ function. As you can see from the appendix, I can't reach $g$ in $\psi$, so I can't build $g$ to behave like a member of the $A$ set.
How can I overcome this problem?
APPENDIX: example of proof with functional reduction
Let's show that $B = \{ x\in \mathbb{N} | W_x = \emptyset \}$ is productive.
We do this by proving that $\overline{K}$ reduces to $B$, so equivalently we can show that $K$ reduces to $\overline{B} = \{x\in \mathbb{N} | W_x \neq \emptyset \}$. Consider the function: $$ \psi(x,y) = \left\{\begin{array}{lr} 1, & \text{for } x \in K\\ \uparrow, & \text{otherwise} \\ \end{array}\right\} $$
(Notice that $\psi$ is computable since $K \in RE$). We can now build $\varphi_{g(x)}(y) = \psi(x,y)$ with $g$ total recursive (s-m-n theorem). At this point:
$x \in K \implies \forall y\in \mathbb{N}. \varphi_{g(x)}(y) \downarrow \implies W_{g(x)} = \mathbb{N} \implies W_{g(x)} \neq \emptyset\implies g(x) \in \overline{B}$.
$x \notin K \implies \forall y\in \mathbb{N}. \varphi_{g(x)}(y) \uparrow \implies W_{g(x)} = \emptyset\implies g(x) \notin \overline{B}$.
So $K$ reduces to $\overline{B}$ thorugh $g$: $B$ is productive. $\square$