If $C_2$ is irrational, then there are infinitely many twin primes?

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This is a natural follow-up after question 3629282.

It is trivial that the irrationality of Brun's constant $B_2\approx1.90216$ implies that there are infinitely many twin primes: $$ B_2 \mbox{ is irrational } ~\Rightarrow~ \mbox{ twin prime conjecture is true.} \tag{1} $$

Interestingly, this answer (now deleted) claimed that something similar is also applicable to the twin prime constant $C_2$: if we can prove the irrationality of the twin prime constant $$ C_2 = \prod_{p > 2} \left(1-\frac{1}{(p-1)^2} \right) = 0.66016\ldots \qquad\mbox{(product over all odd primes } p) $$ then necessarily there are infinitely many twin primes?!

However, the implication $$ C_2 \mbox{ is irrational } ~\Rightarrow~ \mbox{ twin prime conjecture is true (?)} \tag{2} $$ is not at all obvious to me. To put it mildly, $(2)$ is far less obvious than $(1)$ for Brun's constant $B_2$.

Could anyone please sketch the reasoning behind $(2)$ if you do see how it can be done?

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I don't think that's true. This constant merely reflects the conjectured asymptotic density of twin primes. If that conjecture is true, there are infinitely many twin primes, irrespective of whether the constant is rational. And if the conjecture is false, this constant has nothing to do with twin primes. So I don’t see why there should be a connection between the irrationality of this constant and the infinitude of twin primes.