= Solution
With $r=R\sin\chi$, \b[the static metric is]
$$
\boxed{ds^2=-\cos^2\chi\,dt^2+R^2d\chi^2+R^2\sin^2\chi\,d\Omega_2^2.}
$$
The radial coefficient is regular at $\chi=\pi/2$, but the time coefficient vanishes and these coordinates cease to be a chart. Part c supplies smooth horizon-crossing coordinates; the local <Rindler coordinates> also prove the degeneracy is a coordinate one. Extending $\chi$ while keeping the same static $t$ alone is not a regular coordinate extension.
The complete analytic extension is the hyperboloid $-Y_0^2+\sum_{i=1}^4Y_i^2=R^2$. Static coordinates are $Y_0=\sqrt{R^2-r^2}\sinh(t/R)$, $Y_4=\sqrt{R^2-r^2}\cosh(t/R)$, and $(Y_1,Y_2,Y_3)=r\mathbf n$. Global coordinates give
$$
ds^2=-dT^2+R^2\cosh^2(T/R)(d\psi^2+\sin^2\psi\,d\Omega_2^2).
$$
With $\tan\eta=\sinh(T/R)$ this is
$$
ds^2=R^2\sec^2\eta(-d\eta^2+d\psi^2+\sin^2\psi\,d\Omega_2^2),\quad-\pi/2<\eta<\pi/2,\quad0\leq\psi\leq\pi.
$$
\b[The <Penrose diagram> is a rectangle], with spacelike past/future infinity, regular pole lines $\psi=0,\pi$, and radial null rays at $45$ degrees. Observer horizons divide static patches from inaccessible regions; they are not curvature boundaries.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-58-de-sitter.png]
{title=Global de Sitter conformal rectangle and the north and south static patches}
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