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The assertion is false without further global hypotheses. For a concrete non-spherical surface of constant Gaussian curvature one, fix $a=1/2$ and setHere $f>0$, $g^{\prime }>0$ and $f^{\prime 2}+g^{\prime 2}=1$, so this is a connected regular embedded surface of revolution with unit-speed meridian. Its metric is $d{u}^2+f(u{)}^{2}d{v}^2$, whose Gaussian curvature is $-f^{\prime \prime }/f=1$. This formula also follows by multiplying its principal curvatures $-f^{\prime \prime }/g^{\prime }$ and $g^{\prime }/f$.

At $u=0$ these principal curvatures are $a=1/2$ and $1/a=2$. Every point of a radius-one sphere is umbilical with both principal curvature magnitudes one. The displayed surface therefore cannot lie in any such sphere, despite having $K=1$ everywhere.

No. Fix $0<A<1$ and, near $s=0$, letThen $f^{\prime 2}+g^{\prime 2}=1$, so this is a regular unit-speed profile with $f>0$ on a sufficiently small interval. Rotate it around the $z$-axis. For an arc-length surface of revolution, $K=-f^{\prime \prime }/f$, and thereforeAt $s=0$, the formulas from part (a) give the principal curvaturesThey are unequal because $A\ne 1$, so the point is not an umbilical point. Every point of a sphere is umbilical. Consequently no neighbourhood of this point can lie on a sphere, giving a non-spherical surface of constant Gaussian curvature one.