= Solution
Set $z=e^y$. The metric becomes
$$
ds^2=z^{-2}\bigl[dz^2-dt^2+d\mathbf x^2\bigr],\qquad z>0,
$$
so these are <Poincare coordinates on anti-de Sitter spacetime> with unit curvature radius. They cover one patch, not the whole global extension.
On the two-dimensional section with the three transverse $x_i$ suppressed, choose embedding coordinates $X_{-1}=\sec\rho\cos T$, $X_0=\sec\rho\sin T$, $X_d=\tan\rho$. The Poincaré transformation is
$$
\boxed{z=\frac{\cos\rho}{\cos T+\sin\rho},\qquad
t=\frac{\sin T}{\cos T+\sin\rho}.}
$$
Choose the connected component containing $T=\rho=0$. Its denominator is positive, giving
$$
\boxed{-\pi/2<\rho<\pi/2,\qquad |T|<\pi/2+\rho.}
$$
Its two sloping edges are <Poincaré horizons>, null coordinate boundaries at $z\to\infty$. The right vertical edge is the $z\to0$ timelike conformal boundary. Substituting the transformation into $z^{-2}(dz^2-dt^2)$ gives $\sec^2\rho(d\rho^2-dT^2)$, verifying the chart and its null edges.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-75-poincare-patch.png]
{title=The Randall-Sundrum Poincaré patch inside global AdS, with null patch horizons and the reference brane z=1}
The full coordinates in this part have $y\in\mathbb R$, hence the whole shaded $z>0$ patch. If one additionally retains only the Randall–Sundrum bulk $y\ge0$ as in Question4, keep $z\ge1$. The brane $y=0$, or $z=1$, has $\cos T=\cos\rho-\sin\rho$ and is drawn in orange; the retained side contains $T=0$, $\rho<0$. A patch horizon is not a curvature singularity or the timelike conformal boundary. A geodesic can cross it in finite proper time while the patch coordinate $t$ diverges.
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