Proving the weak maximum principle for subharmonic functions Background Information:
I am studying the book: Partial Differential Equations by Walter A. Strauss. I will be trying to adapt the proof given in chapter 6 section 1 to prove the following question.
Definition: 
If $U$ is a bounded, open set, we say that $v\in C^2(U)\cap C(\overline{U})$ is subharmonic if 
$$-\Delta v \leq 0 \ \ \ \text{in} \ U$$
This may be relevant for this particular proof from what I have read online.
Relevant proof to be considered:
a.) Prove that if $v$ is subharmonic and $B(x,r)\subseteq U$, then 
$$v(x) \leq ⨍_{\partial B(x,r)}v(y)dS_y$$
Proof a.) - Let 
$$f(r) = ⨍_{\partial B(x,r)}v(y)d S_y = ⨍_{|y-x| = r}v(y)d S_y = \frac{1}{n\alpha(n)r^{n-1}}\int_{|y-x| = r}v(y) dS_y = \frac{1}{n\alpha(n)}\int_{|z| = 1}u(x+rz) d S_z$$
where $y = x + rz$ and $\alpha(n) = \int_{|x|\leq 1}dx = $ volume of unit ball in $n$ dimensions. The outer unit normal at $y\in\partial B(x,r)$ is $\nu(y) = \frac{y-x}{r}$, thus 
\begin{align*}
f'(r) = \frac{1}{n\alpha(n)}\int_{|z|=1}z \nabla v(x+rz) dS_z &= \frac{1}{n\alpha(n)r^{n-1}}\int_{|y-x| = r}\nabla v(y) \frac{y-x}{r}d S_y\\
&= \frac{1}{n\alpha(n)r^{n-1}}\int_{|y-x| = r}\nabla v(y)\nu(y)d S_y
\end{align*}
Since $-\Delta v\leq 0$ in $U$, we have 
$$0\leq \int_{B(x,r)}\Delta v(y) dy = \int_{\partial B(x,r)}v_{\nu}d S_y$$
thus $f'(r)\geq 0$ for $r > 0$. Then we get for $r > 0$
$$v(x) = f(0) \leq f(r) = ⨍_{\partial B(x,r)}v(y)d S_y$$
Therefore,
$$v(x) \leq ⨍_{\partial B(x,r)}v(y)dS_y$$
b.) Prove that if $v$ is subharmonic and $B(x,r)\subseteq U$, then 
$$v(x)\leq ⨍_{B(x,r)}v(y)dy$$
Proof b.) - 
Following from part a.), using polar coordinates, we get
\begin{align*}
⨍_{B(x,r)}v(y)dy = \frac{1}{\alpha(n)r^{n}}\int_{B(x,r)}v(y)dy &= \frac{1}{\alpha(n)r^{n}}\int_{0}^{r}d\rho \int_{\partial B(x,\rho)}v(y)dS_y\\
&\geq \frac{1}{\alpha(n)r^{n}}\int_{0}^{r}n\alpha(n)\rho^{n-1}v(x)d\rho\\
&= \frac{v(x)}{\alpha(n)r^{n}}\int_{0}^{r}n \rho^{n-1}d\rho\\
&= v(x)
\end{align*}
Hence, 
$$v(x)\leq ⨍_{B(x,r)}v(y)dy$$
Question:

Prove the weak-maximum principle for subharmonic functions: if $v$ is subharmonic, then 
  $$\max_{x\in \bar{U}}v(x) = \max_{x\in \partial\bar{U}}v(x)$$

Attempted proof - Let $\epsilon > 0$ and $v(x) = u(x) + |x|^2$. Then still in two dimensions, say,
$$\Delta v = \Delta u + \epsilon (x^2 + y^2) \geq 0 + 4\epsilon > 0 \ \ \text{in} \ U$$
But $\Delta v = v_{xx} + v_{yy} \leq 0$ at an interior maximum point, by the second derivative test in Calculus. Therefore, $v(x)$ has no interior maximum in $U$. Now, $v(x)$, being a continuous function, has to have a maximum somewhere in the closure $\bar{U} = U\cup \text{bdy} U$. Say the maximum is attained at $x_0\in \text{bdy}U$. Then fore all $x\in U$
$$u(x)\leq v(x) \leq v(x_0) = u(x_0) + \epsilon |x_0|^2\leq \max_{\text{bdy}U} u + \epsilon l^2$$
where $l$ is the greatest distance from $\text{bdy}U$ to the origin. Since this holds for all $\epsilon > 0$ we have $$u(x)\leq \max_{\text{bdy}U}u \ \forall x\in U$$
Attempted proof 2 - Assume that $v$ takes an interior maximum at some point $x_0$. Then for $r > 0$, if $B(x_0,r)\subseteq U$ by part b.) above we have
$$⨍_{B(x_0,r)}v(y) - v(x_0) \geq 0$$
but this integrand is nonpositive so $v(y) = v(x_0)$ for all $y\in B(x_0,r)$
I am not sure where to go from here, any suggestions or comments are greatly appreciated.
 A: 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$.
A: Here is a proof from Vector Calculus, 6th edition, by Mardsen and Tromba where $
\overline U$ is the unit disc. See Exercises 46 and 47 in Section 3.3.
We use the fact that given a sequence $\{\mathbf p_n\}, n = 1,2,\dotsc$, of points in a closed bounded set in $\Bbb R^2$, there exists a point $\mathbf q$ such that every neighborhood of $\mathbf q$ contains at least one member of $\{\mathbf p_n\}$.
If $v_n(x,y) = v(x,y) + e^x/n$, then $\Delta v_n = e^x/n > 0$. Hence, $v_n$ is strictly subharmonic and can have its maximum only on $\partial \overline U$, say, at $\mathbf p_n = (x_n,y_n)$. If $(x_0,y_0) \in \overline U$, it is easy to show that $u(x_n,y_n) > u(x_0,y_0) - e/n$. Hence, there must be a point $\mathbf q = (x_\infty,y_\infty)$ on $\partial \overline U$ such that arbitrarily close to $\mathbf q$ we can find an $(x_n,y_n)$ for $n$ as large as we like. We then conclude from the continuity of $v$ that $v(x_\infty,y_\infty) \ge v(x_0,y_0)$.
