Look for a large gap in the distribution of primes. For this conjecture, the gap between n!+2$n!+2$ and n!+n$n!+n$ will suffice. Set y = n!+2$y = n!+2$ (which is composite) and set m$m$ (which will be (x+y)/2$\frac{x+y}{2}$, so x$x$ will be 2m-y$2m-y$ eventually) to be the largest composite so that there are no primes between m$m$ and y$y$. Then there will be primes between x$x$ and m$m$ and none between m$m$ and y$y$. So pi(x) + pi(y) = pi(x) + pi((x+y)/2) > 2 pi((x+y)/2)
So $\pi(x) + \pi(y) = \pi(x) + \pi\left(\frac{x+y}{2}\right) > 2 \pi\left(\frac{x+y}{2}\right)$. So your conjecture will fail for infinitely many pairs x$x$ and y$y$.
Gerhard "This Belongs On Another Forum" Paseman, 2015.06.30
