$f$ is analytic on $D$, prove that $f$ is a constant

Here is another idea. The intuition is this. We modify $f$ so that $f(0) = 0$ so we can apply Schwarz lemma. Then we want to "Taylor" expand at $z = 1$.

Assume $f$ isn't constant. Let $f_1 = \psi \circ f$ where $\psi$ is the conformal map chosen so $\psi(1) = 1$ and $\psi(f(0)) = 0$. We can take $$ \psi(z) = w \frac{z - f(0)}{1 - \overline{f(0)} z} $$ where $w$ is in the unit circle so $\psi(1) = 1$.

Let $h(r) = f_1(r)$ if $r \in (0,1)$ and $h(r) = 1$ else. Also $h'(r) = \psi'(f(r))f'(r)$ so $h$ is continuously differentiable and $h'(1) = 0$ and $h(1) = 1$. We need here that $|f(0)| < 1$.

Since $f$ isn't constant, by Schwarz lemma $|{h(r)}| < r$ for $r < 1$.

Now $\left| \frac{h(1-\epsilon) - h(1)}{-\epsilon} \right| < \frac 12 $ if $\epsilon$ is small. Hence $|h(1-\epsilon)| \geq 1 - \frac 12 \epsilon \geq 1-\epsilon$.