Constant Gaussian curvature disks

This is an addendum to the proofs by Anton and Deane, completing the missing part of the argument.

Lemma. Let $D$ be the closed unit disk and $f: D\to S^2$ an immersion such that $f(\partial D)$ is a circle $C$ in $S^2$. Then $f$ is 1-1.

Proof. Let $J: S^2\to S^2$ denote the reflection in $C$. Double $D$ cross its boundary to obtain the 2-sphere $\Sigma$ and let $j: \Sigma\to\Sigma$ denote the reflection in $\partial D$. Then extend $f$ to a local homeomorphism $F: \Sigma\to S^2$ by $$ F(j(z))=J f(z). $$ Since $\Sigma$ is compact, $F$ is a covering map, hence, a homeomorphism. Thus, $f$ is 1-1. qed

Edit. As for the existence of the isometric immersion $\iota$ in Anton's answer (above, $\iota=f$), it is an application of Riemann's theorem (the local form of the Killing–Hopf theorem): It shows that for every $z\in D$ there exists neighborhood $U\subset D$ and an isometric embedding $U\to S^2$. Joe Wolf in his book attributes the local result to Riemann and he is probably right, but it is likely (since it is about surfaces) that Gauss already knew how to prove this.

Since $D$ is simply-connected, these local isometries can be combined to produce a globally-defined isometric immersion, see for instance, the answer to this question.