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With my highest respect, I always look at your answers in MathOverflow to learn. Technically the last equality in your second display is wrong, because the integral has an asymptotic $=\frac{n}{\log n}-\frac{45n}{4\log^7n}+O(\log^{-1}n)$, and you retain the lesser error term $O(n^{\frac12+\varepsilon})$ Nevertheless your reasoning is still correct. I have some doubts about this conjecture, by the reason I give in my note below. There have to be some collaboration between the terms $\pi((n+1)^2)-\textrm{Li}((n+1)^2)$ and $\pi(n^2)-\textrm{Li}(n^2)$ for this to be true.
If you substitute each $\pi(\cdot)$ by the corresponding $\textrm{Li}(\cdot)$, the resulting function tends to $1$. But the difference $\pi(x)-\textrm{Li}(x)$ has been shown to be $\Omega_{\pm}(\frac{x^{1/2}\log\log\log x}{\log x})$. Therefore one of the error $\pi(x^2)-\textrm{Li}(x^2)$ can be of order $\frac{x}{\log x}\log\log\log x$. Therefore the sup limit of this divided by $\pi(x)$ is $+\infty$. So only if the error in $\pi((x+1)^2)$ and $\pi(x^2)$ collaborate can the limit be $1$.
In the paper: Hiary, Ghaith A. An explicit van der Corput estimate for ζ(1/2+it). Indag. Math. (N.S.) 27 (2016), no. 2, 524–533. It is proved that $|\zeta(1/2+it)|\le 0.63 t^{1/6}\log t$ for $t\ge 3$ and we have $|\zeta(1/2+it)|\le 1.461$ for $0\le t\le 3$. Of course the proof is more complicate than that in the answer of Carlo Beenakker.
