Let me restate the problem the way I understand it:
we have 2 events A and B separated by a space-like interval
$$\Delta s^2=\Delta x^2 -(c\Delta t)^2 > 0 $$
now, different observers will measure these 2 events A and B and come up with different $\Delta x$ and $\Delta t$, but what will be the minimum possible $\Delta x$ (if it exists) that one of these observers might measure?
from $$ 0\geq-(c\Delta t)^2$$ we obtain $$ \Delta x^2\geq\Delta x^2-(c\Delta t)^2)=\Delta s^2$$
So $\Delta x^2\geq\Delta s^2$ always in a spacelike interval and therefore the minimum value that it can attain is $\Delta s^2$, for an observer that sees A and B happen simultaneously i.e. with $\Delta t=0$
Now let's try a different , more physically insightful, apprach.
First a little trick which will help better visualize the situation: let's agree that all observers reset their clocks, meters,etc such that event A has coordinates (0,0) for every observer. This does not change the motion of an observer and in general the physics of any problem. So A=(0,0) for everybody, while B=(t,x) has different coordinates for different observers, but for everybody $\Delta s^2=x^2-t^2$ is the same, say $\Delta s^2=9$ ($c=1$ from now on). Every observer will draw a space-time diagram with event A at the origin (not shown) and event B appearing somewhere. If we overlay all the diagrams we get the following
where each observer has drawn B as a different colored dot at different positions. All these dots belong to the locus $\Delta s^2=x^2-t^2=9$ so it is clear that the green observer will measure the smallest $\Delta x^2$ and $\Delta t^2=0$
PS: the light cone in the diagram is a bonus, I could not resist putting it in ;-)