I would like to prove this result without any appeals to theorems like, say the three-lines theorem, but only through the use of the fundamental solution, $\log |z - w|$ in the plane.
The proof I have seen goes as follows:
Consider $u : \mathbb{C} \to \mathbb{R}$ a subharmonic, and bounded function. Consider a logarithmic perturbation of $u$, given by: $$ v_\epsilon(z) = u(z) - \epsilon \log |z| $$ Observe that when $|z| > 1$, we have that $\log |z| > 0$, so that $v_\epsilon(z) < u(z)$ here. Moreover (here is the portion of the proof I take issue with): $$ \sup_{|z| > 1} v_\epsilon = \max_{S(0,1)}v_\epsilon = \max_{S(0,1)} u = \max_{\overline{D}(0,1)}u $$ Thus, for any $z$ such that $|z| > 1$, we have that: $$ u(z) = v_\epsilon(z) + \epsilon \log |z| \leq \max_{\overline{D(0,1)}} u + \epsilon \log|z| $$ As the LHS is independent of $\epsilon$, we may send $\epsilon$ to $0$, and conclude that $u(z) \leq \max_{\overline{D(0,1)}}u$, and taking suprema, get that $\sup_{\mathbb{C}}u = \max_{\overline{D(0,1)}}u$. Thus, $u$ attains an interior maximum, and hence is constant.
Question: Why are we allowed to apply the maximum principle with $v_\epsilon$ ? I though that the maximum principle only applied on bounded domains?
Since $u$ is bounded, $v_\epsilon(z) \to - \infty$ as $|z|\to +\infty$. Therefore, the maximum of $v_\epsilon(z)$ must be obtained on the unit circle.