Here is my though process:
- The standard unit sphere is an Alexandrov space with curvature bounded from below by $1$.
- An octant of that sphere (including boundary) is a convex subset of that sphere with respect to the intrinsic metric, so it is an Alexandrov space of curvature bounded from below by $1$.
- If we glue two copies of an octant together along their boundaries (via the obvious isometry), we obtain an Alexandrov space of curvature bounded from below by $1$ (by Doubling theorem). The obtained space is homeomorphic to the sphere.
- A sphere with curvature bounded from below by $1$ can be realized by a closed convex surface in $\mathbb R^3$. Here closed convex surface in $\mathbb R^3$ means boundary of a bounded convex domain. This surface is unique up to isometry. (Due to Alexandrov/Pogorelov).
There might be a flaw in my logic, if not: What does the realization of the double of an octant in $\mathbb R^3$ look like? I'm having trouble imagining it; for a flat a disk or hemisphere it is easy.