the lemma has two parts, but both start from the premise that $f$ is a holomorphic function that goes from $\mathbb{D}=\{z: |z| < 1\}$ to itself with $f(0)=0$:
the proof does make heavy use of the maximum modulus principle
it says that a non-constant $f$ can't attain a maximum on any domain $\Omega$, and if $\exists z_0: |f(z_0)| \ge |f(z)|\ \forall z \in \Omega$, then $f$ is constant.
$f$ has a zero at $0$, so we can write $f(z)=zg(z)$ where $g$ is also some holomorphic function
notice that if we fix an $R \in (0, 1)$, $|z|=R \implies |g(z)| = \left|\frac{f(z)}{z}\right| \le \frac{1}{R}$.
this value of $\frac{1}{R}$ has to be the largest $f$ can achieve for $|z| < R$ as well, since all holomorphic functions have to achieve their maximum on the boundary of a compact set.
if we let $R \to 1$, then we get $|g(z)| \le 1$ for ALL $z$ in $\mathbb{D}$.
that's basically the first part done, since
\[|f(z)| = |z| \cdot |g(z)| \le |z|\]also since $f'(0)=g(0)$ by power/taylor series stuff, $|f'(0)|$ is bounded above by $1$ too.
ok so if $|f(z)|=|z|$ for any $z \ne 0$ then $|g(z)|=1$
same thing applies if $|f'(0)|=1$, since again $f'(0)=g(0)$
i guess the point is that we have a point inside the disc where $g(z)=1$, even though we already bounded it by $1$
$g$ must therefore be constant, and its uniform value is what we take $\lambda$ to be
this one also deals with $f: \mathbb{D} \to \mathbb{D}$, and it gives a straightforward bound:
\[|f'(z)| \le \frac{1-|f(z)|^2}{1-|z|^2}\]also equality at any point implies that $f$ is a mobius transformation, i.e. it's of the form
\[f(z)=e^{i\theta}\frac{\alpha-z}{1-\overline{\alpha}z}\quad \theta \in \R, |\alpha| < 1\]proof is highkey jank- fix any $z_0 \in \mathbb{D}$ and consider these functions:
\[\begin{align*} S(z)=\frac{z+z_0}{1+\overline{z_0}z} && T(z)=\frac{z-f(z_0)}{1-\overline{f(z_0)}z} && F=T \circ f \circ S \end{align*}\]through bashing you'll see that $F(0)=0$, so we can apply the previous version of schwarz's lemma to say that $|F'(0)| \le 1$
but by the chain rule we can also see that
\[\begin{align*} |F'(0)| &= |T'(f(S(0))) \cdot f'(S(0)) \cdot S'(0)| \\ &= |T'(f(z_0)) \cdot f'(z_0) \cdot S'(0)| \\ &\le 1 \end{align*}\]through even more bashing you can obtain the derivatives
\[\begin{align*} S'(z)=\frac{1-|z_0|^2}{(1-\overline{z_0}z)^2} && T'(z)=\frac{1-|f(z_0)|^2}{(1-\overline{f(z_0)}z)^2} \end{align*}\]and so
\[\begin{align*} |f'(z_0)| &\le \frac{1}{|T'(f(z_0))| \cdot |S'(0)|} \\ &= \frac{(1-\overline{f(z_0)}f(z_0))^2}{1-|f(z_0)|^2} \cdot \frac{(1-\overline{z_0} \cdot 0)^2}{1-|z_0|^2} \\ &= \frac{1-|f(z_0)|^2}{1-|z_0|^2} \end{align*}\]the choice of $z_0$ was arbitrary, so this is true for all $z_0 \in \mathbb{D}$
if equality's hit at any point, then for some $z_0$ our construction gets $|F'(0)|=1$ and we can say that $F(z)=\lambda z$ with $\lambda=1$
by EVEN MORE bashing it can be seen that $T$ and $S$ are invertible, so we can write $f$ like so:
\[f=T^{-1} \circ F \circ S^{-1}\]when you compose all these formula together $f$ does indeed take the form of a mobius transform
(i mean cmon get like wolfram alpha to do it or smth idk)