I want to know if my proof is correct. This is what I have to show: Let $u:\mathbb{R}^n\rightarrow \mathbb{R}$ be harmonic and bounded above. Show that $u$ is constant. I want to apply Harnack's inequality for non-negative harmonic functions $$\sup_{B(x_0, r)}u\leq c\inf_{B(x_0, r)} u, $$ when $B(x_0, 4r)\subset \Omega$ and where $c=3^n.$ So let $R>0$ and $x\in B(0, R)$. $u$ is bounded above so $u(y)\leq M$. Define $v=M-u$. Then $v$ is harmonic and $v\geq 0$. By Harnack's inequality we have $$v(x)\leq\sup_{B(0, R)}v\leq c\inf_{B(0, R)}v\leq cv(0).$$ Because $R$ was arbitrary we conclude that $v$ is bounded above. Since it's also bounded below, by Liouville's theorem $v$ is constant, thus $u$ is constant.
Am I right?
You are right.
I feel like I should write some more, so here's an alternate argument by generalising a proof of Liouville's theorem. Consider any two points $x,y$. Let $c$ denote the volume of $B(0,|x-y|)$. The integral of $u$ in $B(x,2|x-y|)$ is $c2^n u(x)$, and the integral of $u$ in $B(y,|x-y|)$ is $c u(y)$. This means that the average value of $u$ in $B(x,2|x-y|)\setminus B(y,|x-y|)$ is
$$\frac{2^n u(x) - u(y)}{2^n - 1} = u(y) + \frac{2^n}{2^n-1}(u(x)-u(y)).$$
In particular there exists a point $x'$ with $u(x')-u(y)\geq \frac{2^n}{2^n-1}(u(x)-u(y))$. So if we start with points $x_0,y$ with $u(x_0)>u(y)$, we can inductively pick points $x_1,x_2,x_3,\dots$ with
$$u(x_k)-u(y)\geq \left(\frac{2^n}{2^n-1}\right)^k(u(x_0)-u(y))\to+\infty.$$
Thus any non-constant harmonic function is unbounded above.