= Solution
\b[The assertion is false without further global hypotheses.] For a concrete <non-spherical surface of constant Gaussian curvature one>, fix $a=1/2$ and set
$$
f(u)=a\cos u,\qquad g(u)=\int_0^u\sqrt{1-a^2\sin^2 t}\,dt,
$$
$$
X(u,v)=(f(u)\cos v,f(u)\sin v,g(u)),\qquad |u|<\pi/4,\quad v\in\mathbb R/(2\pi\mathbb Z).
$$
Here $f>0$, $g'>0$ and $f'^2+g'^2=1$, so this is a connected regular embedded <surface of revolution> with unit-speed meridian. Its metric is $du^2+f(u)^2dv^2$, whose <Gaussian curvature> is $-f''/f=1$. This formula also follows by multiplying its <principal curvatures> $-f''/g'$ and $g'/f$.
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.
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