Proving the weak maximum principle for subharmonic functions

Let us proof by contradiction. We know that $$\max_{x\in \bar{U}}v(x) \ge \max_{x\in \partial\bar{U}}v(x)$$

Suppose that $$\max_{x\in \bar{U}}v(x) > \max_{x\in \partial\bar{U}}v(x) $$

Then $v$ takes an interior maximum at some point $x_0 \in U$. So, we have that there is a point $x_0 \in U$, such that

$$v(x_0) > \max_{x\in \partial\bar{U}}v(x) \tag{1}$$

By part b.) above we have, for any $r > 0$, if $B(x_0,r)\subseteq U$, $$⨍_{ B(x_0,r)}v(y) - v(x_0) = ⨍_{ B(x_0,r)}(v(y) - v(x_0)) \geq 0$$ but the integrand $(v(y) - v(x_0))$ is nonpositive so $v(y) = v(x_0)$ for all $y\in B(x_0,r)$

Let us consider $L = \{r>0 |B(x_0,r) \subseteq U\}$ and $s=\sup L$

Since $B(x_0,s) = \bigcup_{r \in L}B(x_0,r) \subseteq U$, we have that, for all $y\in B(x_0,s)$, $v(y) = v(x_0)$

Since $v \in C(\overline{U})$, we have that, for all $y\in \overline{B(x_0,s)}$, $v(y) = v(x_0)$.

Claim: $ \overline{B(x_0,s)} \cap \partial U \neq \emptyset$.

Once this claim is proved, we immediately have that there is $y \in \partial U$, such that $v(y) = v(x_0)$, which contradicts $(1)$.

Proof of the claim:

Let us proof it by contradicion. Suppose that $ \overline{B(x_0,s)} \cap \partial U = \emptyset$.

Since $ \overline{B(x_0,s)} \subset \overline{U} $, so if $ \overline{B(x_0,s)} \cap \partial U = \emptyset$, then $\overline{B(x_0,s)} \cap (\mathbb{R}^n \setminus U) = \emptyset$.

But since $\overline{B(x_0,s)}$ is compact and $(\mathbb{R}^n \setminus U)$ is closed, if they are disjoint, then

$$ 0< d=\inf \{ |b-a| \; | \; b \in \overline{B(x_0,s)} \textrm{ and } a \in (\mathbb{R}^n \setminus U)\} $$

So $B(x_0,s+\frac{d}{2})$ and $(\mathbb{R}^n \setminus U)$ are disjoint, which means that $B(x_0,s+ \frac{d}{2}) \subseteq U$. So $s+\frac{d}{2} \in L$ and $s+\frac{d}{2} >s =\sup L$. Contradiction. So we must have $ \overline{B(x_0,s)} \cap \partial U \neq \emptyset$.