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 $$\frac1{m(B(p,r))}\int_{B(p,r)}v<M.$$
This follows directly from mean-value property of harmonic functions. We will assume that $\Omega$ is open.
Since $K \subset \Omega$ is compact and $\Omega\subset \mathbb{R}^n$ is open then $r: = \mathrm{dist}(K ,\partial \Omega) > 0$ (notice that if $\Omega = \mathbb{R}^n$, then $r = \infty$, in which case we take $r$ as any positive number).
In view of the definition of $r$, for $x\in K$ we have $\overline{B(x,r - \varepsilon)} \subset \Omega$ for any fixed $\varepsilon > 0$ small enough. In view of the mean value property for $u$ we have
$$
u(x) = \frac{n}{\omega_n (r - \varepsilon)^n} \int\limits_{B(x,r - \varepsilon)} u(y) dy,
$$
hence
$$
|u(x)| \leq \frac{n}{\omega_n (r - \varepsilon)^n} \int\limits_{B(x,r - \varepsilon)} |u(y)| dy \leq \frac{n}{\omega_n (r - \varepsilon)^n} \int\limits_{ \Omega } |u(y)| dy.
$$
The right-hand side of the last inequality does not depend on $x$, and taking $\sup$ on the left-hand side over $x \in K$ we get
$$
\sup\limits_{x \in K} |u(x)| \leq \frac{n}{\omega_n (r - \varepsilon)^n} \int\limits_{ \Omega } |u(y)| dy,
$$
for any $\varepsilon > 0$ small. Taking $\varepsilon \to 0$ produces the desired inequality.
Best Answer
Let $\epsilon > 0$. Then there exists $\delta > 0$ such that $|u(x)-u(y)|<\epsilon$ for all $y\in B(x,\delta)$. So, for $r<\delta$, \begin{align} \left|u(x) - \frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)} u\,\mathrm{d}S\right| &= \left|\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)}(u(x)-u)\,\mathrm{d}S\right|\\ &\le \frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)}|u(x)-u|\,\mathrm{d}S\\ &< \epsilon. \end{align}