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 $M$.
