In classical differential geometry, Liebmann’s theorem states that a compact and connected surface in $\mathbb{R}^{3}$ with constant Gauss curvature is a (standard) sphere; see e.g. Do Carmo's Differential Geometry of Curves and Surfaces (2nd edition, p. 323). It follows that any sphere is rigid.
On the other hand, the sphere is not the only surface in $\mathbb{R}^{3}$ with constant Gauss curvature. For instance, there exist surfaces of revolution with constant Gauss curvature equal to 1 that have no umbilical point. In Kristopher Tapp's Differential Geometry of Curves and Surfaces (p. 222), they are called fake spheres. Here is an image taken from the same book.
What I do not understand is how the surface corresponding to $a=2$ in the picture fails to be smooth, compact (assuming that it contains its boundary) or connected.
I have always thought that the rigidity of the sphere was referring to the sphere as a closed surface (compact surface without boundary), whereas a single hemisphere could be deformed isometrically in space.
So my question is, why don't we need to assume closedness (instead of compactness) in the statement of Liebmann’s theorem?