Why pseudo-Riemannian metric cannot define a topology? It is not clear for me why a positive definite metric is necessary to define a topology as noted in some textbooks like the one by Carroll.
Does this imply that in cosmology, say through FLRW metric, we can only discuss the topology or global geometry of space, or spatial hypersurface, instead of spacetime?
Also related to this question is that we know there "exists" a coordinate system in which the pseudo-Riemannian metric in GR becomes, locally, a Lorentzian one, thus having canonical signature - + + +. 
In FLRW metric we assume an isotropic and homogeneous cosmos based on observation in the, well, "observable" universe.
But how about breaking this assumption and imagine that a global Riemannian metric or coordinates system "exists" for spacetime and only demand that it locally becomes Lorentzian?
Which part of my understanding is correct and which one incorrect?
 A: Usually a differentiable metric define a distance on a manifold which in turns induces the topology, a pseudo-riemannian metric you don't have a distance.
A: You asked for "Why pseudo-Riemannian metric cannot define a topology?", but they can. 
For traditional metric $d$, a ball as the building block of an open set is defined as $B(x,\epsilon)=\{y\mid d(x,y)<\epsilon\}$. This definition gives a set of sets with no regard on how $d$ is defined. This set of sets gives a sub-basis of a topology. (Just as any set of subsets from the original space can be declared as a sub-basis, yielding the smallest topology that contains all those sets.)
This is very probably not the answer that you want to see, but I hope it will help to improve your general understanding so that you can pose a question more specific to what you have in mind.
Edit: probably the best that you can get without losing contact to physics is https://en.wikipedia.org/w/index.php?title=Pseudometric_space&oldid=683593022#Metric_identification. By this, you collapse each world-line that has $v=c$ into a single "point". I guess this can be justified from a physics point of view as no proper time is elapsed while traveling along these lines.
A: I cannot personally answer that question but I can make a list of books or internet sites where I saw related discussions:
Remark 1: A space-time is usually defined as a Lorentzian manifold, and a manifold is a topological space. So we assume that our space-time already has a topology. Nevertheless, there is still the question if one can associated another topology based on the pseudo-metric and whether it agrees with the original topology.
Remark 2: the assumption that there exists a Riemannian metric is void since that is always the case, or at least I remember a statement of this kind.


*

*The analogue of balls of radius r,  $\left\lbrace q ∈ V_+(p), d(q,p) < r \right\rbrace$ in a Lorentzian metric fail to form a basis of the topology... (Ok I have to check a few things): here $V_+(p)$ is the future/forward directed (causal?) cone and $d$ is some kind of pseudo-distance obtained from the pseudo-metric... or maybe this is wrong after all, cf. this question

*4.24. p.34, Techniques of Differential Topology in Relativity, Roger Penrose: the Alexandrov topology coincides with the natural topology of the manifold  iff this latter is strongly causal, but it is coarser in general.

*Remark 2.4 p.374, General relativity and the Einstein equations, Yvonne Choquet-Bruhat: one can associate a Riemannian metric to a time-oriented Lorentzian manifold. Then one can define balls, and thus a basis of a topology.

*6.1 p.378, General relativity and the Einstein equations, Yvonne Choquet-Bruhat


I'll improve the answer later...
