= Solution
For a continuously differentiable curve $\gamma(t)=x(t)+iy(t)$ in the <Poincaré half-plane model>, its hyperbolic length is
$$
\boxed{L_H(\gamma)=\int_a^b
\frac{\sqrt{\dot x(t)^2+\dot y(t)^2}}{y(t)}\,dt}.
$$
The hyperbolic lines are the <Geodesic in the Poincare half-plane model>[geodesics]: vertical Euclidean lines and Euclidean semicircles whose centres lie on the real axis.
The unique hyperbolic line through distinct $z,w\in H$ can be carried by an upper-half-plane <Möbius transformation> to the imaginary axis. Such transformations are isometries by hypothesis, so it is enough to consider endpoints $iy_0$ and $iy_1$. Every joining curve satisfies
$$
L_H(\gamma)
\ge\int_a^b\frac{|\dot y|}{y}\,dt
\ge\left|\log\frac{y_1}{y_0}\right|.
$$
Equality holds for the monotone vertical parametrization
$$
\gamma(t)=iy_0\exp\!\left(t\log\frac{y_1}{y_0}\right),
\qquad 0\le t\le1.
$$
Thus the appropriately parametrized hyperbolic-line segment $[z,w]$ attains the infimum defining $\rho(z,w)$.
Now let $l,m$ have positive infimum distance. They neither intersect nor share an ideal endpoint: an intersection would give distance zero directly, while lines sharing an ideal endpoint approach one another arbitrarily closely near that endpoint. Hence they are ultraparallel. An upper-half-plane isometry puts them into the form
$$
l=\{re^{i\theta}:0<\theta<\pi\},
\qquad
m=\{Re^{i\theta}:0<\theta<\pi\},
\qquad 0<r<R.
$$
This is the normal form underlying the <common perpendicular of ultraparallel hyperbolic lines>. Put $z=e^{u+i\theta}$. Since
$$
dx^2+dy^2=e^{2u}(du^2+d\theta^2),
\qquad
y=e^u\sin\theta,
$$
the metric becomes
$$
ds^2=\frac{du^2+d\theta^2}{\sin^2\theta}.
$$
Every path from $l$ to $m$ therefore has
$$
L_H\ge\int|du|\ge\log\frac Rr.
$$
Equality requires $\theta=\pi/2$ and monotone $u$, so it is attained on the imaginary axis, which meets both semicircles orthogonally. Its intersection points $ir$ and $iR$ realize the distance
$$
\boxed{d=\log(R/r)}.
$$
For a nonattained example, take the vertical lines $l=\{iy:y>0\}$ and $m=\{1+iy:y>0\}$. They share the ideal endpoint at infinity. The horizontal segment at height $y$ has hyperbolic length $1/y$, so their infimum distance is zero, but distinct points of $H$ always have positive distance. Hence the infimum is not attained.
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