I'll expand on Nelson's proof a bit.
You know the Mean Value Property that the value of a harmonic function $u$ at $z$ equals the average of its values on any circle centered at $z$. We can turn this into a fact about averages over balls (which in this case are disks) just by using polar coordinates:
$$\int_{B(z,R)} u(w)\,dw = \int_0^R \int_0^{2\pi} u(z + r e^{i \theta}) r \,d\theta\,dr = \int_0^R 2 \pi u(z) r\,dr = u(z) \pi R^2.$$
Dividing both sides by $\pi R^2$, we see that $u(z)$ equals the average value of $u$ over the ball $B(z,R)$.
Now fix $z,w \in \mathbb{C}$, and consider disks $B(z,R)$, $B(w,R)$ centered at each of them. Let $L = B(z,R) \cap B(w,R)$ be the lens-shaped region lying in both disks. Let $L_z = B(z,r) \backslash L$ be the region in the $z$ disk but not the $w$ disk. (Draw a picture.) One can work out the area $A(L)$ of $L$ by a bit of multivariable calculus (or look it up on Mathworld). A bit more calculus shows that the ratio $\frac{A(L)}{\pi R^2} \to 1$ as $R \to \infty$, so that the lens occupies "most" of the disk. Since $A(L) + A(L_z) = \pi R^2$, we must therefore have $\frac{A(L_z)}{\pi R^2} \to 0$.
Now by the mean value property, for any $R$ we have
$$u(z) = \frac{1}{\pi R^2} \int_{B(z,R)} u(\zeta)\,d\zeta = \frac{1}{\pi R^2} \int_L u(\zeta)\,d\zeta + \frac{1}{\pi R^2} \int_{L_z} u(\zeta)\,d\zeta.$$
For the second term, if $|u|$ is bounded by some constant $M$, we have as $R \to \infty$,
$$\left|\frac{1}{\pi R^2} \int_{L_z} u(\zeta)\,d\zeta \right| \le \frac{M A(L_z)}{\pi R^2} \to 0.$$
Thus
$$u(z) = \lim_{R \to \infty} \frac{1}{\pi R^2} \int_L u(\zeta)\,d\zeta.$$
But if we replace $z$ with $w$ we can make exactly the same argument (replacing $L_z$ by $L_w$ whose area is the same, by symmetry). So we also have
$$u(w) = \lim_{R \to \infty} \frac{1}{\pi R^2} \int_L u(\zeta)\,d\zeta.$$
Thus $u(z) = u(w)$.
I don't see how you could use Liouville's theorem to prove that, but it does follow from Cauchy-Riemann's equations.
If you assume that $f$ is entire, use Cauchy-Riemann's equation on $|f|^2 = u^2 + v^2$ to show that both $u$ and $v$ must be constant on $D$. After that it follows from the uniqueness theorem that $f$ is constant everywhere.
Best Answer
Suppose $\operatorname{re} f(z) \ge \alpha$ and let $g(z) = {1 \over f(z)-\alpha+1 }$. Since $\operatorname{re} (f(z)-\alpha+1) \ge 1$, we have $|g(z)| \le 1$ for all $z$ hence $g$ is a (non zero) constant. Since $f(z) = \alpha-1+{1 \over g(z)}$, we see that $f$ is constant.
If $\operatorname{re} f(z) \le \alpha$, then repeat the above with $-f$ (and $-\alpha$, of course).