If $x,y$ are time-like and linearly independent, then the vector $x+ty$ is light-like for exactly two values of $t\in\R$.
We may assume $\la y,y\ra=-1$, then $\la x+ty,x+ty\ra=0$ is equivalent to
$$
t^2-2t\la x,y\ra+\Vert x\Vert^2=0
\quad\mbox{i.e.}\quad
t=\la x,y\ra\pm\sqrt{\la x,y\ra^2-\Vert x\Vert^2}
$$
and by the wrong Cauchy-Schwarz inequality the discriminant is strictly positive. This also works in the reverse direction: If $x,y$ are time-like and linearly independent, then the line $x+\R y$ must intersect the light cone, because it doesn't pass through the origin. Hence the discriminant of the quadratic equation $\la x+ty,x+ty\ra=0$ must be non-negative, which in turn implies the wrong Cauchy-Schwarz inequality.