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I think the following argument settles the conjecture that $\inf |\rho_{\zeta}-\rho_{\zeta'}|=0$, assuming RH. Basically, it follows because the zeros can be arbitrarily close.

Littlewood proved that $$\inf \gamma_n-\gamma_{n-1}=0.$$ On RH this implies that $$\inf \rho_n-\rho_{n-1}=0.$$ So, for every $\epsilon>0$ one can find a pair of zeros between which $$\max\{|\zeta(1/2+it)|:\gamma_{n(\epsilon)-1}<t<\gamma_{n(\epsilon)}\}=\epsilon.$$

By the theorem of Macdonald (cited in my question above) and the inverse mapping theorem, the derivative must vanish somewhere on the boundary of the closed curve enclosing $\rho_{n(\epsilon)}$ defined by the condition that $|\zeta(s)|=\epsilon$, and not inside it. By the maximum principle, as $\epsilon\rightarrow 0$ the modulus on and throughout the interior of the curve tends to zero, so such a curve must tend to a point as $\epsilon\rightarrow 0$. Since the derivative vanishes on the boundary, the result follows.

I think the following argument settles the conjecture that $\inf |\rho_{\zeta}-\rho_{\zeta'}|=0$, assuming RH. Basically, it follows because the zeros can be arbitrarily close.

Littlewood proved that $$\inf \gamma_n-\gamma_{n-1}=0.$$ On RH this implies that $$\inf \rho_n-\rho_{n-1}=0.$$ So, for every $\epsilon>0$ one can find a pair of zeros between which $$\max\{|\zeta(1/2+it)|:\gamma_{n(\epsilon)-1}<t<\gamma_{n(\epsilon)}\}=\epsilon.$$

By the theorem of Macdonald (cited in my question above) and the inverse mapping theorem, the derivative must vanish somewhere on the boundary of the closed curve enclosing $\rho_{n(\epsilon)}$ defined by the condition that $|\zeta(s)|=\epsilon$, and not inside it. By the maximum principle, as $\epsilon\rightarrow 0$ the modulus on and throughout the interior of the curve tends to zero, so such a curve must tend to a point as $\epsilon\rightarrow 0$. Since the derivative vanishes on the boundary, the result follows.

EDIT. This argument is not quite right. Actually the curves described above are not necessarily closed, and therefore need not tend to a point. However, the derivative certainly vanishes on the boundary of the largest closed curve enclosing $\rho$ and on which $|\zeta(s)|=c$, and $c$ is less than or equal to $\epsilon$. The conclusion is the same, because these curves are closed.

I think the following argument settles the conjecture that $\inf |\rho_{\zeta}-\rho_{\zeta'}|=0$, assuming RH. Basically, it follows because the zeros can be arbitrarily close.

Littlewood proved that $$\inf \gamma_n-\gamma_{n-1}=0.$$ On RH this implies that $$\inf \rho_n-\rho_{n-1}=0.$$ So, for every $\epsilon>0$ one can find a pair of zeros between which $$\max\{|\zeta(1/2+it)|:\gamma_{n(\epsilon)-1}<t<\gamma_{n(\epsilon)}\}=\epsilon.$$

By the theorem of Macdonald (cited in my question above) and the inverse mapping theorem, the derivative must vanish somewhere on the boundary of the closed curve enclosing $\rho_{n(\epsilon)}$ defined by the condition that $|\zeta(s)|=\epsilon$, and not inside it. By the maximum principle, as $\epsilon\rightarrow 0$, such a curve must also become small and, since the derivative vanishes somewhere on it, the result follows.

I think the following argument settles the conjecture that $\inf |\rho_{\zeta}-\rho_{\zeta'}|=0$, assuming RH. Basically, it follows because the zeros can be arbitrarily close.

Littlewood proved that $$\inf \gamma_n-\gamma_{n-1}=0.$$ On RH this implies that $$\inf \rho_n-\rho_{n-1}=0.$$ So, for every $\epsilon>0$ one can find a pair of zeros between which $$\max\{|\zeta(1/2+it)|:\gamma_{n(\epsilon)-1}<t<\gamma_{n(\epsilon)}\}=\epsilon.$$

By the theorem of Macdonald (cited in my question above) and the inverse mapping theorem, the derivative must vanish somewhere on the boundary of the closed curve enclosing $\rho_{n(\epsilon)}$ defined by the condition that $|\zeta(s)|=\epsilon$, and not inside it. By the maximum principle, as $\epsilon\rightarrow 0$ the modulus on and throughout the interior of the curve tends to zero, so such a curve must tend to a point as $\epsilon\rightarrow 0$. Since the derivative vanishes on the boundary, the result follows.

I think the following argument settles the conjecture that $\inf |\rho_{\zeta}-\rho_{\zeta'}|=0$, assuming RH.

Littlewood proved that $$\inf \gamma_n-\gamma_{n-1}=0.$$ On RH this implies that $$\inf \rho_n-\rho_{n-1}=0.$$ So, for every $\epsilon>0$ one can find a pair of zeros between which $$\max\{|\zeta(1/2+it)|:\gamma_{n(\epsilon)-1}<t<\gamma_{n(\epsilon)}\}=\epsilon.$$

By the theorem of Macdonald (cited in my question above) and the inverse mapping theorem, the derivative must vanish somewhere on the boundary of the closed curve enclosing $\rho_{n(\epsilon)}$ defined by the condition that $|\zeta(s)|=\epsilon$, and not inside it. By the maximum principle, as $\epsilon\rightarrow 0$, such a curve must also become small and, since the derivative vanishes somewhere on it, the result follows.

I think the following argument settles the conjecture that $\inf |\rho_{\zeta}-\rho_{\zeta'}|=0$, assuming RH. Basically, it follows because the zeros can be arbitrarily close.

Littlewood proved that $$\inf \gamma_n-\gamma_{n-1}=0.$$ On RH this implies that $$\inf \rho_n-\rho_{n-1}=0.$$ So, for every $\epsilon>0$ one can find a pair of zeros between which $$\max\{|\zeta(1/2+it)|:\gamma_{n(\epsilon)-1}<t<\gamma_{n(\epsilon)}\}=\epsilon.$$

By the theorem of Macdonald (cited in my question above) and the inverse mapping theorem, the derivative must vanish somewhere on the boundary of the closed curve enclosing $\rho_{n(\epsilon)}$ defined by the condition that $|\zeta(s)|=\epsilon$, and not inside it. By the maximum principle, as $\epsilon\rightarrow 0$, such a curve must also become small and, since the derivative vanishes somewhere on it, the result follows.