When deriving the speed of light in vacuum, one usually starts from Maxwell's equations, does some calculus and finds a wave solution with the phase velocity $c = 1/\sqrt{\mu_0 \varepsilon_0}$. This is clear to me.
When analyzing transmission lines, the approach is not quite the same. One starts with a model of distributed inductance, capacitance, resistance and admittance along the transmission line from which one obtains the telegrapher's equations.
Solving these equations (with a sinusoidal ansatz) again leads to a wave solution of the form $\exp(\mathrm i \omega t - \gamma x)$ where $$ \gamma = \sqrt{(R + \mathrm i \omega L) (G + \mathrm i \omega C)} . $$ For a lossless transmission line ($R = 0 = G$) we thus find $v_{\mathrm{ph}} = \omega / \operatorname{Im}(\gamma) = 1 / \sqrt{L C}$.
Now, in the literature I've been looking at (on calculating $v_{\mathrm{ph}}$ for coplanar waveguides, but this is unimportant, I suspect) it is stated as an obvious fact that in the absence of any dielectric except vacuum $v_{\mathrm{ph}} = c$ in the transmission line. This is not obvious to me at all. Shouldn't the waveguide geometry (which determines $L$ and $C$, after all) be able to affect this value? The idea seems to be that only the space in which the fields propagate determine this velocity and that the conductors, carrying merely currents and charges, can't affect it.
Can you justify this fact and make it intuitive for me?
Does this still hold with a lossy transmission line? That is, if I "switch on" the resistance $R > 0$ without changing the geometry of an existing transmission line or any nearby dielectrics, will $L$ and $C$ "magically" adjust in order to keep $\operatorname{Im}(\gamma)$ (and thus $v_{\mathrm{ph}}$) constant?