If my computations are correct, I do not think the result depends on Riemann hypothesis. Really my computation does not have assumed it. I have not used my gamma's are real.

This is well known. If the sets are independents on a probability space there is no such subsequence.

@Ralph @fedja I have upvoted the solution. I do not see anything illegal. Or do you think anything not relevant to RH is illegal? Of course the example have too many zeros, and with very large multiplicity. I think the solution must be accepted and deserves the bounty. And still I would like to see a solution with the number of zeros rho = beta + i gamma with -T < t < T bounded by C T log T. Perhaps in the line.

If I would add one condition, I would add that the function have of the order of T*log T zeros on the line sigma = 1/2 in the segment 0 < t < T. But even without this I do not know any solution. I would like to see the ones "shameless cheating all the way through"

@Ralph Furmaniak. But I expect that any disk with center at 5/4+i T with radius 1 and T big enough contains points with zeta(s)=1. You have not explained the criptic form of your constant $e^{e^{e^3}}$. I would like to know the reason of this election. This said I am not so sure that there is a function satisfying your conditions. All my examples ends with sup beta = sigma limit where the function is not bounded if beta + i gamma are the zeros. Of course when one is in mood of being skeptical.

$e^{e^{e^3}}\approx 2\times 10^{229\,520\,860}$

Also $\zeta(s)-(s-1)^{-1}$ have zeros on $\sigma>1/2$, example: one near 0.50000132378059385590 + 99997.458357552926821 I. We must recongize that until now we can not give any example of an entire (or even meromorphic with a finite number of poles) function satisfying conditions (1), (2), (3) and (4).