Can anyone one show how speeds greater than c cause a paradox on a space time diagram? So according to Relativity, time moves at different rates for different observers, and distant observers will actually disagree on the order of events.  Now from what I gather this gets sorted out for objects with relative speeds less than c, but would ultimately cause a paradox for anything moving faster because it would time travel.  However, I don't really see how.  On a spacetime diagram light always moves at a 45° angle, or at least some angle between 0° and 90°, and the only way I see anything effecting the past is if something moves at an angle greater than 90°.

So can someone please give a visual representation that explains the issue?
Suppose there was a magic gun that fired a bullet at ten times the speed of light relative to the firer. 

If I have the only such gun, and I don't move then there is no paradox.

But now suppose these guns are generally available.

At time zero, an observer flies past me at velocity 0.5 c and I give them one of these guns. 7 years later, I fire the gun after them, so in my frame of reference the bullet will catch them at about time  7.36 years when they are about 3.68 ly away. 

To get these numbers I just need to solve $10*(t-7) = 0.5*t$ to calculate when the bullet overtakes the spaceship.

In their frame of reference though, that occurs about 
time 6.38 years after they pass me, at which point, I am about 3.19 ly away from them (we both agree on our relative velocity).

This is calculated using the Lorentz transformations. $$t' = \gamma(t - vx)$$ taking $c = 1$. Here $t = 7.38$, $\gamma = \frac{1}{\sqrt{1-v^2}} \sim 1.15$ $v = 0.5$ and $x = 3.68$.

They now fire back and we can do a similar calculation, but in their frame of reference. So in their frame the bullet will pass me at time $t_1$ when 
$(t_1 - 6.38)*10 = 0.5 t_1$ which tells us that $t_1 = 6.71$.

Now we use the Lorentz transformation again to determine what time that is in our frame. We find that it is 5.81 years. A bit more than a year before we fired our bullet.

The key thing is that $n$ times the speed of light in one frame will be backwards in time in another frame.

The diagrams below illustrates this scenario.



Both diagrams show exactly the same events. The one on the left uses the time and space coordinates of the original observer ("you") the one on the right uses the coordinates of the other observer. In bith cases time is on the y axis, space on the x axis. Units are years and light years.

The black diagonal lines are some possible world lines of light rays.
The red lines are lines of constant x (solid) or t (dashed) as determined in "your" frame of reference. The thick red arrow (at x = 0) is your world-line. 

The blue lines are lines of constant x (solid) or t (dashed) as determined by the ship. The thick blue arrow is the ship's world-line. As can be seen you meet at the event that you both agree is x = 0, t= 0.

The green line is the bullet that "you" fire. It crosses three solid red lines and goes a bit less than half way from one dashed red line to the next (travels 3+ ly in one 0.36 y). The cyan line is the return bullet. It crosses 3 solid blue lines and goes rather less than half way from one dashed blue line to the next (travels 3+ ly in about 0.3y). 

Assuming I've got this right, all the intersections and crossing patterns of red, blue and black lines, and the four world-lines should agree between the two diagrams, since they really do depict the same events.  In particular both diagrams show a causal loop, in which the cyan bullet arrives back at the red observer before the green bullet is fired. They disagree about which bullet travels back in time, but that's relativity for you. 

What this tells us practically is that any physics which allows FTL travel but not time travel must not be Lorentz invariant, so your gun will work differently in different frames of reference. We know of no such physics at present, and see no clues which suggest it might exist.