We first claim that if a $C^0(\bar{\Omega})$ function $u$ satisfies the following local mean value inequality
$$u(y)\leq\frac{1}{n\omega_nR^{n-1}}\int_{\partial B_R(y)}u ds \text{ for all } R\leq \delta$$
then maximal principle applies, i.e. $u$ attains its maximum at $\partial \Omega$.
Assume our claim now. Let $\tilde{u}$ be the harmonic function defined by the boundary value $u|_{\partial B}$ on the boundary of an arbitrary ball $B\subset\subset \Omega$. Then $\tilde{u}$ satisfies mean value equality in $B$ since it is harmonic in $B$. Then $u-\tilde{u}$ satisfies the above mean value inequality. Then by our claim (maximal principle), we have $\sup_{x\in B}(u-\tilde{u})=\sup_{x\in\partial B}(u-\tilde{u})$. But $u-\tilde{u}=0$ on $\partial B$. Hence we have $u-\tilde{u}\leq 0$. $u\leq \tilde{u}$ in $B$.
Now we show that $u$ is subharmonic, given an harmonic function $h$ in $B\subset \subset \Omega$, such that $u\leq h$ on $\partial B$. We need to show that $u\leq h$ in B. Since we know $u\leq\tilde{u}$ in $B$. It suffices to show that $\tilde{u}\leq h$ in $B$. To see this one uses the Possion kernel $$\tilde{u}(x)=\int_{\partial B}K(x,y)u(y)ds_y$$
$$h(x)=\int_{\partial B}K(x,y)h(y)ds_y$$
Since $u(y)\leq h(y)$ on $\partial B$ and $K(x,y)\geq 0$. We have $\tilde{u}(x)\leq h(x)$.
Now we show our claim that the local mean value inequality
$u(y)\leq\frac{1}{n\omega_nR^{n-1}}\int_{\partial B_R(y)}u ds $
implies maximal principle.
We use the standard topological argument. Define a set $S:=\{q\in\bar{\Omega}| u(q)=\sup_{x\in\bar{\Omega}}u(x)\}$. If $u$ attains maximum at a point $p\in\Omega$ then $S$ is nonempty since $p\in S$. Note that $S$ is closed since every sequence in $S$ converges to a point in $S$ since $u$ is continuous. Next if $p\in S\cap\Omega$, then $u(p)\leq\frac{1}{n\omega_nR^{n-1}}\int_{\partial B_R(p)}u ds $ and $u(p)\geq u(x)$ implies $u(x)=u(p)$ for all $x\in B_R(p)$ since $u$ is continuous. Hence $B_R(p)\in S$
Hence $S$ is open. Now $S$ is open and closed and nonempty, hence we have $S=\bar{\Omega}$. This proves our claim and finished our proof.
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
You can find a (sketch) of the proof on Wikipedia, based on approximation by convolutions. A different approach, based on the solvability of the Dirichlet problem for the laplacian, can be read in Axler's book here.