Space of univalent mappings $f: \mathbb{D} \to \mathbb{C}$ has no nesting

Let $$S$$ be the space of univalent (i.e. injective) mappings from the disk $$\mathbb{D}$$ to the plane $$\mathbb{C}$$ normalized so that $$f(0) =0$$ and $$f'(0)=1$$. So

$$f(z) = z+a_2z^2+a_3z^3+\cdots.$$

I'd like to show that if $$f,g \in S$$ and $$f(\mathbb{D}) = g(\mathbb{D})$$ then $$f=g$$.

Attempt: I'm aware the Schwarz lemma implies this. Let $$U = f(\mathbb{D})= g(\mathbb{D})$$ and let $$F: U \to \mathbb{D}$$ be a Riemann conformal mapping with $$F(0)=0$$. Then the Schwarz lemma applies to $$F\circ f$$ and $$F\circ g$$. But other than seeing that $$F'(0)\leq 1$$, I do not see what this gets me. Any help would be appreciated.

If $$f(\mathbb{D}) = g(\mathbb{D})$$ then $$h = g^{-1} \circ f$$ is a holomorphic function from $$\mathbb{D}$$ into $$\mathbb{D}$$ with $$h(0) = 0$$ and $$h'(0) = 1$$. Now use the Schwarz Lemma to conclude that $$h$$ is the identity.