In relativity (both special and general) one of the key quantities is the proper length given by:
$$ ds^2 = g_{\alpha\beta}dx^\alpha dx^\beta \tag{1} $$
where $g_{\alpha\beta}$ is the metric tensor. The physical significance of this is that if we have a small displacement in spacetime $(dx^0, dx^1, dx^2, dx^3)$ then $ds$ is the total distance moved. You can think of it as a spacetime equivalent of Pythagoras' theorem. The quantity $ds$ is an invariant i.e. all observers in any frame of reference will agree on the value of $ds$.
A metric is positive definite if $ds^2$ is always positive, and Riemannian manifolds have a metric that is positive definite.
However in relativity $ds^2$ can be positive, zero or negative, which correspond to timelike, lightlike and spacelike intervals respectively. It is because $ds^2$ can have different signs that manifolds in GR are not Riemannian but only pseudo-Riemannian.
Lorentzian manifolds are a special case of pseudo-Riemannian manifolds where the signature of the metric is $(3,1)$ (or $(1,3)$ depending on your sign convention).
If we take the metric tensor that corresponds to special relativity (i.e. flat spacetime) equation (1) becomes:
$$ ds^2 = -dt^2 + dx^2 + dy^2 + dz^2 $$
That minus sign on the $dt^2$ term means $ds^2$ can be negative as well as positive, so the manifold is pseudo-Riemannian, and the one negative and three positive signs on the right hand side make the signature $(3,1)$ so the manifold is Lorentzian.
It is simply false, at least written as it stands.
The point is that the relation between the topology and the metric is more complicated than in the Riemannian case, where the geodesical balls form a basis of the topology$^1$.
As a matter of fact, a (connected) Lorentzian smooth metric $g$ over the time-oriented smooth manifold $M$ does define a topology, the same already present on $M$ if the spacetime is strongly causal.
Fix a point $p\in M$ and consider all (smooth) timelike future-directed curves through $p$ and denote by $L(\gamma)$ the Lorentzian length of $\gamma= \gamma(\xi)$, $\xi \in [a,b]$.
$$L(\gamma) = \int_a^b \sqrt{|g(\dot{\gamma},\dot{\gamma})|} d \xi$$
If $q \in M$ define the so called Lorentzian distance of $q$ from $p$ as
$$\tau(q,p) := \sup \{L(\gamma) \:|\: \mbox{$\gamma$ timelike future-directed from $p$ to $q$}\}$$
If no timelike future-directed from $p$ to $q$ exists, $\tau(q,p) :=0$.
Next define $I^+(p) := \{q \in M \:|\: \tau(q,p)>0\}$ and $I^-(p) := \{q \in M \:|\: \tau(p,q)>0\}$.
It is possible to prove that the family of sets $I(p,q):= I^+(p) \cap I^-(q)$ (with the suitable chronological order of the arguments) is a basis of the topology of $M$ if the spacetime is strongly causal [Kronheimer and Penrose (1967)]. (Strongly causal means that every open neighborhood $U$ of every event $p\in M$ includes another open neighborhood $V$ of $p$ such that $J^+(r) \cap J^-(s) \subset V$ if $r,s \in V$, Minkwski spacetime and all globally hyperbolic spacetimes like Kruskal's one are strongly causal.)
This is a well known result of semi-Riemannian geometry (Theorem 4.9 in Global Lorentzian Geometry second edition 1996 by J.K. Beem, P.E. Ehrlich, K.L. Easley)
ADDENDUM. To answer to a comment to my answer, in view of the quoted result, the topology induced by sets $I(p,q)$ is metrizable since it coincides with the natural topology of the manifold $M$ viewed as a smooth manifold regardless any (semi)-Riemannian structure thereon, which is metrizable. In particular, if $M$ is Minkowski spacetime a distance producing the said topology can be constructed explicitly: $$d((t,\vec{x}), (t',\vec{x}')) = ||\vec{x}-\vec{x}'||+c|t-t'|$$
The balls of this distance are evidently the sets $I((t,\vec{x}),(t',\vec{x}'))$. This distance evidently depends on the choice or the Minkowkian reference frame.
(1) In a connected Riemannian manifold $M$ whose metric is denoted by $g$, $d(p,q) = \inf \{ L(\gamma) \:|\: \gamma \mbox{ smooth curve joining $p$ and $q$}\}$ where
$$L(\gamma) := \int_a^b \sqrt{g(\dot{\gamma},\dot{\gamma})} d \xi$$
is a distance making $M$ a metrical space. All the open balls $B_\delta(p):=\{ q\in M \:|\: d(p,q)<\delta\}$, varying $p\in M$ and $\delta \in (0,+\infty)$, form a basis of the topology already present in $M$.
Best Answer
$(n{-}1){+}1$ dimensional de Sitter space can be constructed as a sphere of spacelike radius in $n{+}1$ dimensional Minkowski space. The sign of the squared radius breaks the symmetry.
The sign of the scalar curvature is ambiguous for mixed-signature spaces, since the Ricci scalar inherits the sign convention of the metric. It's probably better to say that de Sitter space has spacelike curvature (spacelike radius of curvature).
Anti de Sitter spaces are also maximally symmetric and can be constructed as spheres of timelike radius in a flat space of $(n{-}1){+}2$ signature. They are topologically $\def\R{\mathbb R} \R^{n-1}\times S^1$ (or $\R^n$ if you take the universal cover, as is typically done so that time isn't cyclic).
The flat spaces $\R^{p+q}$ are also maximally symmetric.
That probably generalizes to: a sphere in flat $p{+}(q{+}1)$ space whose squared radius has the same sign as the $q{+}1$ dimensions is topologically $\R^p S^q$, and moreover, those and $\R^{p+q}$ are probably the only maximally symmetric $p{+}q$ dimensional manifolds up to covers and quotients by $\mathbb Z_2$. But it's not obvious to me that it's true, and I haven't found a source online that confirms it in a quick search.
The answer to your first question is no, since, for instance, $1{+}1$ dimensional de Sitter space isn't homeomorphic to its universal cover or its quotient by $\mathbb Z_2$.