Uniform interval bound from optimal Selberg weights ($\pi (a+x)-\pi (a)\le (2+o(1))x/\log x$)

For all interval origins $a\ge 0$, the number of multiples of any $d$ in $(a,a+x]$ is $x/d+O(1)$ with an absolute error bound. The Selberg sieve diagonalization minimizes ${\sum }_{d,e\le z}{\lambda }_d{\lambda }_e/[d,e]$, subject to ${\lambda }_1=1$, at $1/G(z)$. The optimal Selberg weights have modulus at most one, so the total remainder is $O(z^2)$. Restoring the at most $z$ small primes gives $\pi (a+x)-\pi (a)\le z+x/G(z)+O(z^2)$. Take $z=\lfloor \sqrt{x}/(\log x{)}^2\rfloor$ and use the Selberg sieve denominator asymptotic. The resulting $o(1)$ is independent of $a$.