 Solution

Use the prior itself as an independence proposal: draw $J\ge 1$ with $\mathbb{P}(J=j)=2^{-j}$ and propose $s=2^{-J}$. This indeed samples $p(s)=s$, since the geometric series sums to one. Starting from any allowed current value $r$, accept the proposal with probabilityThe Hastings proposal ratio cancels the prior ratio in the posterior, leaving exactly the binomial likelihood ratio; its binomial coefficient also cancels. All allowed states communicate, and proposing the current state gives a self-loop. Detailed balance therefore gives the specified posterior as the stationary law, and the chain converges to it. Its normalizing constant is finite and positive because it is a prior average of a nonnegative, nonzero likelihood.