By rotational symmetry we may assume that $Du(x_0)$ points in the direction of the first basis vector $e_1$. We must prove that
$${\int\!\!\!\!\!\!-}_{\partial B_R} [u-u(x_0)]\nu_1\le \sup u-u(x_0)$$
where $\nu_1$ is the first component of the unit normal vector. By the mean value property $u-u(x_0)$ has zero mean on $\partial B_R$. Thus, we can add any number to $\nu_1$ without changing the integral. Let's add $1$:
$${\int\!\!\!\!\!\!-}_{\partial B_R} [u-u(x_0)]\nu_1={\int\!\!\!\!\!\!-}_{\partial B_R} [u-u(x_0)](\nu_1+1)$$
Now that the factor $\nu_1+1$ is nonnegative, we use a one-sided bound on $u-u(x_0)$:
$${\int\!\!\!\!\!\!-}_{\partial B_R} [u-u(x_0)](\nu_1+1)\le (\sup u-u(x_0)){\int\!\!\!\!\!\!-}_{\partial B_R}(\nu_1+1)$$
Finally, $${\int\!\!\!\!\!\!-}_{\partial B_R}(\nu_1+1)=1$$ because $\nu_1$ has zero mean.
Well first we should note that the result you state is not true without some smoothness assumptions on $u$. Let's assume $u$ is continuous. We also assume $u$ is real-valued.
There's a standard proof of the result for surface averages that doesn't involve the things you're concerned about and that works just as well for volume
averages:
Suppose $\overline{B(0,1)}\subset\Omega$. We will show that $u$ is harmonic in $B=B(0,1)$. Let $\phi$ be the restriction of $u$ to the boundary of the unit ball, and let $$f=P[\phi],$$the Poisson integral of $\phi$. Then $f$ is harmonic in $B$ and extends continuously to $\overline B$, with values on the boundary given by $\phi=u$. We will let $f$ denote the extension of the original $f$ to $\overline B$. Now define $v:\overline B\to\mathbb R$ by $$v(x)=u(x)-f(x)\quad(x\in\overline B).$$If we can show $v=0$ in $B$ we're done, since $f$ is harmonic in $B$.
Suppose that $v$ does not vanish identically in $B$. Wlog assume that $v$ is strictly positive at some point of $B$. Define $$M=\sup_{x\in B}v(x)$$ and $$K=\{x\in B\,:\,v(x)=M\}.$$
Since $v$ is continuous in $\overline B$ and vanishes on the boundary it follows that $K$ is a nonempty compact subset of $B$. Let $p$ be a point of $K$ at minimal distance to the boundary of $B$.
Now $v$ satisfies the volume mean-value property in $B$ since both $u$ and $f$ do. So for small $r>0$ we have $$M=v(p)=\frac1{m(B(p,r))}\int_{B(p,r)}v.$$But this is impossible: $v\le M$ everywhere in $B(p,r)$ and $v<M$ in a nonempty open subset of $B(p,r)$, so $$\frac1{m(B(p,r))}\int_{B(p,r)}v<M.$$
Best Answer
Hint: consider $\displaystyle\lim_{r\to 0}g(r)$.