The Schwarzschild-deSitter spacetime in Euclidean signature is given by:
$$ ds_{E}^{2}=\left(1-\frac{2M}{r}-\frac{H^{2}r^{2}}{3}\right)d\tau^{2}+\frac{dr^{2}}{\left(1-\frac{2M}{r}-\frac{H^{2}r^{2}}{3}\right)}+r^{2}d\Omega_{2}^{2} $$ where $\tau$ is the Euclidean time, r is the radial coordinate, M is the mass of the black hole and $\Lambda=H^{2}$ is the cosmological constant.
If we set $H=0$ we retrieve the familiar Schwarzschild metric which is asymptotically flat. The latter has only one horizon given by $r=2GM/c^{2}$. Since the horizon singularity is a coordinate artifact, when going to Euclidean time and near the Rindler wedge we can define a periodicity of $\tau$ such that the horizon is regular.
However, when $H\neq0$ the spacetime has two horizons, the black hole and the de Sitter. In this case Hawking and Bousso showed that the metric cannot be made regular in Euclidean signature since one cannot match the periodicity of $\tau$ near both horizons, except for the degenerate case in which they coincide.
The question is: Will an observer crossing either of the two horizon experience a singularity? I am confused about this because the curvature tensor is not singular at any of the horizons.