Constant Gaussian curvature disks This question has also been posted on MSE, but maybe here is the right place to post it.
Is it true that if $D$ is a Riemannian $2$-disk having constant Gaussian curvature equal to $1$ and whose boundary has constant geodesic curvature, then $D$ is isometric to some geodesic ball of the unit sphere $\mathbb{S}^2 \subset \mathbb{R}^3$? I strongly suspect so, but I couldn't find a reasonable argument.
 A: 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.
A: The answer is yes.
Since curvature is 1, there is an isometric immersion $\iota\colon D\looparrowright \mathbb{S}^2$. Note that the curve $\iota(\partial D)$ has constant curvature, therefore $\iota(\partial D)$ bounds a round disc $\Delta\subset\mathbb{S}^2$.
Acually we have two choices for $\Delta$, but for the right choice we get an isometry $D\to \Delta$.
The latter can be done by using Morse-type argument for a function $f\colon\Delta\to\mathbb{R}$.
