= Solution
In the <Poincare disc model>, <hyperbolic lines> are diameters and circular arcs meeting the boundary circle orthogonally. In the <upper half-plane model>, they are vertical lines and semicircles with centres on the real axis. The metric is conformal, so hyperbolic perpendicularity agrees with Euclidean perpendicularity.
Use an <isometry> of the <hyperbolic plane> to send an arbitrary line to the positive imaginary axis. For $P=x+iy$ with $x\ne0$, a point on that axis is $Q=iv$ with $v>0$. The <hyperbolic distance> formula gives
$$
\cosh\rho(P,iv)=1+\frac{x^2+(y-v)^2}{2yv}=\frac{x^2+y^2+v^2}{2yv}.
$$
Put $R=\sqrt{x^2+y^2}$. The last expression is $(v+R^2/v)/(2y)$, uniquely minimized at $v=R$, since $v+R^2/v\geq2R$ with equality only there. Hence the unique closest point is $Q_0=iR$. The semicircle $|z|=R$ passes through $P$ and $Q_0$ and meets the imaginary axis at a right angle. Conversely a semicircle centred at a real number $b$ can meet that axis orthogonally only if $b=0$; a vertical line cannot furnish such a crossing. Thus the perpendicular is unique. Isometries transfer both uniqueness and minimization back to the original line.
For $P$ on the specified second semicircle, $a-r\leq x\leq a+r$ and $R=|P|\leq a+r$. Its perpendicular to the axis is the semicircle just found, with radius $R$. Moreover
$$
\cosh d(P,L_1)=\frac{R}{y},\qquad \tanh d(P,L_1)=\frac{x}{R}.
$$
Since $t\leq\operatorname{artanh}t$ for $0\leq t<1$, we obtain
$$
\boxed{d(P,L_1)\geq\frac{x}{R}\geq\frac{a-r}{a+r}>0.}
$$
This is uniform over the whole semicircle, including points arbitrarily close to the ideal endpoints.
Finally send any first line of an ultraparallel pair to the axis by an <isometry> of the <hyperbolic plane>. The second cannot be a vertical line, since that would share the ideal point at infinity. Its two finite endpoints lie on the same side of zero; otherwise its semicircle would meet the axis, or share the endpoint zero. Reflect horizontally if necessary, so those endpoints are $a-r,a+r$ with $a>r>0$. The uniform estimate above then gives
$$
\boxed{d(L_1,L_2)\geq\frac{a-r}{a+r}>0.}
$$
Thus <ultraparallel hyperbolic lines> have strictly positive separation; disjointness in the interior alone would not suffice.
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