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 $(1)$.
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.
Best Answer
Jose's answer gives details of Nelson's original reasoning, but they aren't quite the same as the details in the proof on Wikipedia.
The crucial point in the latter is that we assume, without loss of generality, that $f$ is a nonnegative function (we can assume this because we assumed $f$ is bounded from above or below). Then nonnegativity is used in the first displayed inequality to say that an integral over $B_r(y)$ must be at least as large as the integral over $B_R(x)$, since the latter is a subset of the former.