Possible independence of a generic sentence? Let the generic sentence $P :$  "$\exists z \in Z \text{ such that }p(z) \text{ is true }$".
In addition $Z$ is recursively enumerable, and  for a given $z_0$ in $Z$,  "$p(z_0)$ or $ \lnot p(z_0)$" is  provable.  
If we find $z_0$ such that $p(z_0)$ is true, then the proof of $p(z_0)$ is also a proof of $P$.      
If $P$ is independent (i.e. neither provable nor refutable, for a given system of axioms), then in particular there is no proof of $P$.  So in particular we can't find $z_0$ such that $p(z_0)$ is true (because else there is a proof of $P$).     Then $P$ is false (because $p(z_0)$ is not independent), contradiction.  

Question: Is it a correct proof that $P$ can't be  independent ?

I don't think so, I think there is a beginner mistake but I don't understand where is it.   
Example: Let $Z$ be the set of all the non-trivial zeros of the Riemann zeta function, and $p(z)$ the proposition "$Re(z)  \neq 1/2$". For a given $z_0$ in $Z$, the sentence "$p(z_0)$ or $ \lnot p(z_0)$" is provable.
So if there is not the beginner mistake I'm talking about, then the Riemann hypothesis is not independent.    
The purpose of this example is to show that there is certainly the beginner mistake I'm talking about, because there is no (yet) a proof that the Riemann hypothesis is not independent (see the MO post: Can the Riemann hypothesis be independent?), so that it's very unlikely that the trivial proof above is correct.
 A: In order to use a familiar theory, we use $\mathbb{N}$ instead of $Z$. 
We need to distinguisish between completeness and decidability. A theory $T$ is complete if for any sentence $\varphi$ of the appropriate language, one of $\varphi$ or $\lnot\varphi$ is a theorem of $T$. A theory $T$ (over a countable language) is decidable if there is an algorithm that, given any sentence $\varphi$, will (ultimately) halt and say yes if $\varphi$ is a theorem of $T$, and will ultimately halt and say no if $\varphi$ is not a theorem of $T$. 
Let $T$ be first-order Peano Arithmetic. It turns out that (if the theory is consistent) then $T$ is neither complete nor decidable. The usual proof of the incompleteness of $T$ uses the undecidability of $T$, but the notions are logically distinct. 
Your question seems to be about Incompleteness, and not about Undecidability.
Let $\varphi(x)$ be a formula with $x$ as the only variable that has a free occurrence in $\varphi$. Then for every natural number $n$, either $\varphi(\bar{n})$ is true in $\mathbb{N}$, or false in $\mathbb{N}$. Here $\bar{n}$ is the formal object $SS\dots S\bar{0}$. But it is convenient to use $n$ instead.
Your argument goes as follows. Suppose that for every $n_0$, one of $\varphi(n_0)$ or $\lnot\varphi(n_0)$ is a theorem of PA.  Suppose moreover that there is no $n_0$ such that $\varphi(n_0)$. Then $\exists z\varphi(z)$ is false in $\mathbb{N}$. That is correct. 
Then it looks as if you conclude that $\lnot\exists \varphi(z)$ is a theorem of PA. So it looks as if you are concluding (partial) Completeness of PA, for this restricted class of sentences.
That part is not true. If PA is consistent, then there are formulas $\varphi(x)$ of the type you describe such that $\lnot\exists x\varphi(x)$ is not a theorem of PA.
Such formulas can even be constructed from suitably encoded versions of the assertion that the Diophantine equation $P(x_1,\cdots,x_n)=0$ has a solution in the  natural numbers, where $P$ is an explicit, albeit messy, polynomial with integer coefficients.    
