The shortest interval for which the prime number theorem holds 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.)
 A: I'm not an analytic number theorist, so take this all with many grains of salt. 
Let $S(x,y,p)$ be the set of integers in the interval $(x,x+y)$ which are NOT divisible by $p$. The argument you are imagining for the prime number theorem is
$$\pi(x+y)-\pi(x) = \left| \bigcap_{p \leq \sqrt{x}} S(x,y,p) \right| \approx y \prod_{p \leq \sqrt{x}} \frac{|S(x,y,p)|}{y} \approx y \prod_{p \leq \sqrt{x}} \left( 1 - \frac{1}{p} \right).$$
Your point is that the second $\approx$ is not good on a term by term basis if $p>y$, because the true value of $|S(x,y,p)|/y$ will be either $1$ or $1-1/y$, not $1-1/p$. You therefore suggest that the whole composite approximation should also not be good. I see two immediate issues:
A product of inequalities is not an inequality While it is true that $|S(x,y,p)|/y$ is not $1-1/p$, this formula can be wrong in either direction. So it is possible that the errors cancel and the products are close to equal.
The sketched proof doesn't work for large $y$ Even when $y$ is as large as $x$, there is a huge issue: $\prod_{p \leq \sqrt{x}} (1-1/p) \approx e^{- \gamma}/\log \sqrt{x} = 2 e^{-\gamma}/\log x$, not $1/\log x$ as we want. So there is already something sketchy here. I don't have an intuition for why the right constant in the PNT is $1$, not $e^{-\gamma}$ or $2 e^{- \gamma}$, but since I already know that there is an issue with this sort of argument, I wouldn't take it too seriously in predicting exactly when PNT would fail. (To clarify, I know many arguments why the constant must be $1$: For example, $\sum_{p \leq n} \frac{n}{p} \log p$ should be $\approx \log n! \approx n \log n$, and this only works if the constant is $1$. And I know why the sieve argument doesn't rigorously prove that $\pi(x) \approx e^{-\gamma} x / \log x$. What I don't have is a gut level understanding of why the sieve formula is right up to the constant factor, but not actually right.)
None of this amounts to an argument FOR the conjectures of Granville and Soundararajan, it just argues that I wouldn't take your heuristic particularly seriously. 
A: The short answer is no.  It is likely to be sufficient, although right now the best known is
actually that pi(x + x^{0.525}) > pi(x) for all large x (and likely all x > 117). If it were necessary,
this would go against the conjectures of Granville and Soundararajan that you cite, as epsilon
smaller than 1/2 would not be good enough.
