= Solution
Differentiating the supplied coordinates gives
$$
dX=\cosh(\alpha t)dx+\alpha x\sinh(\alpha t)dt,\qquad
c\,dT=\sinh(\alpha t)dx+\alpha x\cosh(\alpha t)dt.
$$
Subtract the squares. The cross terms cancel and $\cosh^2-\sinh^2=1$, yielding
$$
\boxed{ds^2=c^2dT^2-dX^2.}
$$
The spacetime is flat <Minkowski spacetime> in <Rindler coordinates>. $x>0$ covers $X>|cT|$, the right <Rindler wedge>, while $x<0$ covers the left wedge $X<-|cT|$; neither is the whole spacetime.
The null combinations are
$$
X+cT=x e^{\alpha t},\qquad X-cT=x e^{-\alpha t}.
$$
For $x=C e^{-\alpha t}$ the first is constant; for $x=C e^{\alpha t}$ the second is constant. Thus part (a)'s exponentials are simply the inertial light rays $X\pm cT=\text{constant}$. Their coordinate-time asymptotes reflect the chart's <Rindler horizon>, not a failure of light to move at $c$ in an inertial frame.
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