Consider the two-dimensional case, i.e. of the hyperbolic place. Define our hyperboloid as the set of points $x=(x_1,x_2,x_3)$ in 3-space (note: Minkowski space, but not needed for this problem) that fulfill $x_1^2 - y_1^2 - z_1^2 = 1$ and $x_1>0$. A hyperbolic line is defined as the intersection of a plane that goes through the origin with the hyperboloid. Two distinct hyperbolic points (points on the hyperboloid) determine a hyperbolic line. Consider the drawing labelled  on this picture for clarification (from this website):
I would like to prove something that seems intuitively clear:
Given a hyperbolic line $h$ and a hyperbolic point $p$ not on $h$, there are infinitely many hyperbolic lines through $p$ that do not intersect $h$.
(note: this is the hyperbolic variant of Euclid's parallel postulate)
My attempt at a proof begins like this:
We have by definition that $h$ is the intersection of the plane $m_1$ determined by two distinct hyperbolic points and the origin with the hyperboloid. As $p$ is not contained by $h$, we have that $p$ cannot be contained by $m_1$ either, so any hyperbolic line spanned by $p$ is the intersection of a different plane $m_2$, determined by $p$, the origin and some point $s$, with the hyperboloid. $m_1$ and $m_2$ are thus distinct planes but because they meet at least at one point, the origin, they cannot be parallel and hence we know from Euclidean geometry that their intersection is a line $l$. Because $h$ is on the hyperboloid, if $l$ and $h$ intersect it means that the hyperbolic line spanned by $p$ intersects $h$ in a hyperbolic point, so we want to show that there are infinite ways to choose the point $s$ so that $h$ and $l$ do not intersect.
and at this point I was going to use a topological argument to show that we can select infinitely more points $s$ so that $h$ and $l$ do not intersect than not. But I could not do this, I think my method might be off and that algebra is a better way to go. Does anyone here know how to show this or can point me to a reference with a proof?