If $v$ is subharmonic in the complex plane $\Bbb C$ then
$$ \tag 1
v(z) \le \frac{\log r_2 - \log |z|}{\log r_2 - \log r_1} M(r_1, v)
+ \frac{\log |z| - \log r_1}{\log r_2 - \log r_1} M(r_2, v)
$$
for $0 < r_1 < |z| < r_2$, where
$$
M(r, v) := \max \{ v(z) : |z| = r \} \quad .
$$
That is the "Hadamard three-circle theorem" for subharmonic
functions, and follows from the fact that the right-hand side of
$(1)$ is a harmonic function which dominates $v$ on the boundary
of the annulus $\{ z : r_1 < |z| < r_2 \}$ .
(Remark: It follows from $(1)$ that $M(r, v)$ is a convex function of $\log r$.)
Now assume that $v(z) \le K$ for all $z \in \Bbb C$.
Then $M(r_2, v) \le K$, and $r_2 \to \infty$ in the inequality $(1)$
gives
$$ \tag 2
v(z) \le M(r_1, v)
$$
for $0 < r_1 < |z|$. It follows that
$$
v(z) \le \limsup_{r_1 \to 0} M(r_1, v) = v(0)
$$
because $v$ is upper semi-continuous. Thus $v$ has a maximum
at $z=0$ and therefore is constant.
Remark: As noted in the comments, the condition “$v$ is bounded above”
can be relaxed to
$$\liminf_{r \to \infty} \frac{M(r, v)}{\log r} = 0 $$
which is
still sufficient to conclude $(2)$ from $(1)$.
If $u$ has a local maximum at $x_0 \in \Omega$, that means that there is an open neighbourhood $U\subset \Omega$ of $x_0$ such that $u\lvert_U$ has a global maximum at $x_0$. Since clearly $u\lvert_U$ is harmonic on $U$, by the form of the maximum principle that you already know, it follows that $u\lvert_U$ is constant.
The two harmonic functions $u$ and $v \colon z \mapsto u(x_0)$ hence coincide on a nonempty open set (namely on $U$), and since $\Omega$ is connected, and harmonic functions are real-analytic, the following identity theorem (the case $n = 2$ with the usual identification $\mathbb{C}\cong \mathbb{R}^2$) implies that $u \equiv v$ on $\Omega$, i.e. $u$ is constant.
Identity theorem: Let $W\subset \mathbb{R}^n$ a connected open set, and $f,g \colon W \to \mathbb{R}$ real-analytic. If there is a nonempty open set $V\subset W$ such that $f\lvert_V \equiv g\lvert_V$, then $f \equiv g$ on all of $W$.
Proof: It suffices to consider $g \equiv 0$, since $f\equiv g \iff f-g \equiv 0$. Consider the set
$$Z = \{ x \in W : \text{there is a neighbourhood } U \text{ of } x \text{ such that } f\lvert_U \equiv 0\}.$$
By its definition, $Z$ is an open subset of $W$ (if $x\in W$ and $U$ is a neighbourhood of $x$ such that $f\lvert_U \equiv 0$, then $y\in Z$ for all $y$ in the interior of $U$). Since $f$ is real analytic, we can also describe $Z$ as
$$Z = \{ x \in W : D^{\alpha}f(x) = 0 \text{ for all } \alpha \in \mathbb{N}^n\},$$
for the power series representation of $f$ about $x$,
$$f(y) = \sum_{\alpha \in \mathbb{N}^n} \frac{D^{\alpha}f(x)}{\alpha!}(y-x)^{\alpha},$$
vanishes in a neighbourhood of $x$ if and only if all coefficients vanish. By continuity of $D^{\alpha}f$, the set
$$Z_{\alpha} = (D^{\alpha}f)^{-1}(0)$$
is closed (in $W$) for every $\alpha \in \mathbb{N}^n$, and hence so is the intersection
$$Z = \bigcap_{\alpha \in \mathbb{N}^n} Z_{\alpha}.$$
Thus $Z$ is an open and closed subset of $W$, and since $W$ is connected, we have either $Z = W$ or $Z = \varnothing$. But $Z \neq \varnothing$ was part of the hypotheses of the theorem, so $Z = W$ and indeed $f \equiv 0$ on $W$.
Best Answer
A very very slick proof due to Edward Nelson follows from the version of mean value that integrates over thr whole disc rather than just the circle (can be obtained by integrating the standard mean value property). Pick two points and write the value of the function as integrals over the two discs both of the same radius. Let their radius go to infinity, the symmetric difference of the two discs gets smaller and smaller in proportion to their overlap. Since the function is bounded, the average of the function on one disc is then essentially the average of the function on the intersection of the discs. Hence as the radius goes to infinity, the average over either disc goes to the same number and the value of the function is the same at both points.