 Solution

Choose a very ample divisor $A$ such that $A-D$ has a nonzero global section; this is possible by taking a sufficiently high ample twist. Since $X$ is integral, multiplication by its $m$th power injects $H^0(X,mD)$ into $H^0(X,mA)$. By Serre vanishing and the Hilbert polynomial, $h^0(X,mA)=O(m^n)$. ConsequentlyThis is the polynomial bound for sections of a fixed divisor. The same proof works for a coherent sheaf of support dimension $d$, by choosing the multiplying section to avoid its associated points; the bound is then $O(m^d)$.