It is well known that the prime number theorem on the form \begin{align*} \pi(x+y) - \pi(x) \sim \frac{y}{\log (x+y)} \end{align*} breaks down for short enough intervals, e.g. taking $y=(\log x)^\lambda$ for any $\lambda>1$, as shown by Maier. As to what is short enough (or long enough), both Granville, p. 7 and Soundararajan, p. 79 conjecture that the prime number theorem holds for all $x$ as long as $y\geq x^\epsilon$. However, I believe to have a heuristic argument for the following conjecture:
Conjecture The choice of $y= \sqrt{x}$ is necessary and sufficient for \begin{align*} \pi(x + y)-\pi(x) \sim \frac{y}{\log (x+y)} \end{align*} to hold for all $x$ as $x\rightarrow \infty$.
I understand that this can be considered a bold claim, so my question is:
Q: Considering the heuristic below, is the conjecture as stated reasonable?
Heuristic: To understand why this conjecture should hold, we need to look at the short intervals between consecutive primes squared, defined by $s_k:=\{p_{k}^2, \dots p_{k+1}^2-1\}$ for $k\geq 1$. These intervals naturally occur in the context of the sieve of Eratosthenes, and in particular, each $s_k$ has the specific quality of being fully sieved by the $k$ first primes; any element in $s_k$ is either divisible by some $p \in \mathcal{P}_k:=\{p_1, \dots,p_{k}\}$ or else is a prime $p \notin \mathcal P_k$. In addition, the exact distribution of primes in $s_k$ is in its entirety build up of the periodic sequences \begin{align*} \rho_{k}(n):=\begin{cases} p_k & \text{if } p_k \mid n,\\ 1 & \text{otherwise}, \end{cases} \end{align*} which we visualise for the specific example of $s_3$ by the following table: \begin{matrix} n & 25 & 26 & 27 & 28 & \bf{29} & 30 & \bf{31} & 32 & 33 & 34 & 35 & 36 & \bf{37} & \cdots & 48\\ \hline \\ \rho_1(n) & 1 & p_1 & 1 & p_1 & 1 & p_1 & 1 & p_1 & 1 & p_1 & 1 & p_1 & 1 & \cdots & p_1\\ \rho_2(n) &1 & 1 & p_2 & 1 & 1 & p_2 & 1 & 1 & p_2 & 1 & 1 & p_2 & 1 & \cdots & p_2\\ \rho_3(n) &p_3 & 1 & 1 & 1 & 1 & p_3 & 1 & 1 & 1 & 1 & p_3 & 1 & 1 & \cdots & 1 \end{matrix}
Observe that the lengths of the intervals $s_k$ are of the form $|s_k|=2 p_{k+1} g_k-g_k^2$, where $g_k:=p_{k+1}-p_k$, and hence lie on the curves $2 \sqrt{x} g - g^2$, with $g=2n$, $n\geq 1$.
Necessary part As $k\rightarrow \infty,$ Any interval growing slower than $\sqrt{x}$ will eventually be infinitesimal compared to arbitrarily many primes smaller than $p_k$, and cannot be expected to accurately sample the distribution of primes in $s_k$, which derives from the underlying periodic sequences $\rho_j(n)$, $1\leq j \leq k$, and where the largest period is $p_k$. What this suggests is that $y=\sqrt{x}$ is the sharp barrier below which the prime number theorem breaks down.
Sufficient part On the other hand, any interval growing faster than $\sqrt{x}$ will eventually cover arbitrarily many intervals $s_k, s_{k+1}, \dots, s_{m}$. But this results in an underestimate of $\pi(x + y)-\pi(x)$, since $y/\log (x+y)$ assumes constant density of primes across $[x,x+y]$, equal to the density in the final interval $s_m$ covered, hence suggesting that $y= \sqrt{x}$ is also sufficient. At its most extreme, the latter argument is exemplified by the estimate $\pi(x) \sim x/\log x$, which is well known to be an inferior guess of the number of primes up to $x$ compared to $\pi(x) \sim \textrm{li}(x)$.
(I should add that the heuristic argument is presented in greater detail in a draft manuscript I recently added to arXiv, titled Primes in the intervals between primes squared).
ADDED: CLARIFICATION OF HEURISTIC ARGUMENT In an attempt to make the heuristic clearer, consider the table above. If we move across this with intervals smaller than $p_3=5$, there will be some places where we underestimate the number of primes and some places where we overestimate. This effect magnifies for larger $k$ and intervals growing slower than $\sqrt{x}$, and suggests the necessary part of the heuristic. (Consider even measuring the density across $\rho_3(n)$ only. It should be even more obvious then.)
