Since $\left(\frac{1+z}{1-z}\right)^2$ is holomorphic in $\mathbb{D}$, its imaginary part is harmonic, and we have $$\underset{r \uparrow 1}{\lim}v(re^{i\theta})=0 \quad \forall ~ \theta \in [0, 2\pi)$$
$v$ is not a constant function, why is this not a contradiction?
I don't think we can argue that $v$ is not holomorphic on $\overline{\mathbb{D}}$ b/c for a circle $\delta$ close to the unit circle, the maximum of $v$ is $\epsilon$ close to $0$.
Let's compute. Up to computational errors,
$$v(r e^{i \theta}) = \frac{4r (1-r^2) \sin (\theta)}{(1-2r\cos (\theta)+r^2)^2}.$$
For $\theta$ close to $0$ and $r$ close to $1$, this is:
$$v(r e^{i \theta}) = \frac{4r (1-r^2) (\theta+O (\theta^3))}{(1-2r (1-\theta^2/2+O (\theta^4))+r^2)^2} = \frac{4r (1-r^2) (\theta+O (\theta^3))}{((1-r)^2+ r\theta^2+O (\theta^4))^2} = \frac{4r (1-r^2) (\theta+O (\theta^3))}{(1-r)^4+ 2r(1-r)^2 \theta^2+O (\theta^4)}.$$
If you take $1-r \gg \theta \gg (1-r)^3$, then this is roughly:
$$\frac{4r (1-r^2) \theta}{(1-r)^4} = \frac{4r (1+r) \theta}{(1-r)^3} \to + \infty.$$
Hence, the maximum of $v$ on $S(0,1-\varepsilon)$ goes to infinity as $\varepsilon$ goes to $0$ (contrary to what you wrote in your post). There is not contradiction with the maximum principle. And for $\varepsilon = 0$, there is no contradiction either, since you cannot apply the maximum principle (the function is not defined at $1$).