I think what this means is that:
- if two distinct events $a$ and $b$ are separated by a timelike or null future-directed curve then all observers agree on this (not just inertial ones);
- and in this case there is no such curve which also goes from $b$ to $a$ (which all observers also agree on);
- and if there is no such curve as that specified in (1), then all observers agree on this as well.
Note I have not defined what I mean by 'future-directed' but this can be made precise (see below for a slightly rough hack at this).
And no, this is not proven: it's possible to construct spacetimes containing closed timelike curves for which at least (2) fails. Whether such spacetimes are physically plausible is a different question: I personally suspect strongly that they are not, but I think that is also an open question.
Future- and past-directed curves. Consider a point of spacetime $P$ and its tangent space $T_PM$ ($M$ is spacetime here) and let the metric have signature $-2$ (equivalently, $(+,-,-,-)$). Consider timelike vectors $\in T_PM$: for such a vector $\vec{v}$, $g(\vec{v},\vec{v}) > 0$ (this is the definition of being timelike), and equivalently for some basis $\{\vec{e}_i\}$, $v^iv^jg_{ij} > 0$. Now consider a pair of timelike vectors, $\vec{v}, \vec{u}$, and consider $g(\vec{v},\vec{u}) = v^iu^jg_{ij}$.
Now it turns out that this can have either sign. But (this is slightly more subtle), for any three vectors $\vec{v}, \vec{u}, \vec{w}$, if $g(\vec{v},\vec{u}) > 0$ and $g(\vec{u},\vec{w}) > 0$ then $g(\vec{v},\vec{w}) > 0$. This means that vectors form equivalence classes, and we can say that $\vec{v} \sim \vec{u}$ if $g(\vec{v},\vec{u}) > 0$.
And now it turns out that there are just two such equivalence classes (I have not shown this, but it's not that hard). We call one class future-directed, and one class past-directed.
Now a timelike curve in spacetime is future-directed (respectively past-directed) if its tangent vector is everywhere future-directed (respectively past-directed).
Note I have talked about timelike vectors and curves and not for timelike or null vectors and curves, while above I made a claim about timelike or null curves. I think the generalisation is fairly straightforward: you have to change $>$ to $\ge$, and then you need to add some conditions about the vectors not being zero (because clearly $g(\vec{0},\vec{0}) = 0$). However I am not confident enough about this to claim it without looking things up, and I am a bit lazy.
Note: although I once understood this stuff my memory of it is now quite distant: I'd welcome correction.