Let $f$ be a conformal map from the unit disk $\Delta$ onto itself. Let $a=f(0)$ Let $g_a(z) = \frac{z-a}{1-\overline{a}z}$ Then $g_a \circ f$ is a conformal map from $\Delta$ onto itself, with $g_a \circ f(0)=0$ Therefore, by Schwarz's Lemma for all $z \in \Delta$ $|g_a \circ f(z)| \le |z|$

Because $f$ is a conformal map onto $\Delta$ $f^{-1}$ is also a conformal map of $\Delta$ onto itself. $(g_a \circ f)^{-1}(0)=0$ so that by Schwarz's Lemma $|(g_a \circ f)^{-1}(w)| \le |w|$ for all $w \in \Delta$ Writing $w=g_a \circ f(z)$ this becomes $|z| \le |g_a \circ f(z)|$

Therefore, for all $z \in \Delta$ $|g_a \circ f(z)| = |z|$ By Schwarz's Lemma, $g_a \circ f$ is a rotation. Write $g_a \circ f(z) = e^{i \theta} z$ or $f(z) = e^{i \theta} g_a^{-1}$

Therefore, $f$ is a Möbius Transformation.