You need to go much deeper into the physics in order to understand what is really going on. The equation of the wave inside a conductor can be derived from Maxwell’s equations
$$\mathbf{E} = \mathbf{E}_0 e^{j(\alpha z -\omega t)} e^{-\beta z},$$
where \$z\$ is the direction perpendicular to the conductor in our case and
$$\alpha = \omega \sqrt{\mu\varepsilon} \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{1 + \frac{\sigma^2}{\omega^2\varepsilon^2}}}.$$
Here the first exponential is the wave and the second exponential is the attenuating factor. It can be shown that for a bad conductor, satisfying \$\sigma \ll \omega \varepsilon\$, the wave number is
$$\alpha \approx \omega \sqrt{\mu \varepsilon}.$$
It can also be shown that for a good conductor, satisfying \$\sigma \gg \omega \varepsilon\$, the wave number is
$$\alpha \approx \sqrt{\frac{\omega \mu \sigma}{2}} = \sqrt{\mu\varepsilon} \sqrt{\frac{\omega \sigma}{2\varepsilon}}.$$
Hence the speed of propagation for a bad conductor is
$$v_\mathrm{p} = \frac{\omega}{\alpha} \approx \frac{1}{\sqrt{\mu\varepsilon}}.$$
This is the same as the speed of light in non-conductive media. However in a good conductor
$$v_\mathrm{p} = \frac{\omega}{\alpha} \approx \frac{1}{\sqrt{\mu\varepsilon}} \sqrt{\frac{2\omega\varepsilon}{\sigma}}.$$
It can be seen that the speed of light you would expect to see is multiplied by a second factor, which is less than one due the fact \$\sigma \gg \omega \varepsilon\$. Thus we arrive at the counter intuitive result that the better the conductor the lower the propagation speed.
But wait, there’s more! We use conductors only as anchors for the wave. The wave propagates in the space around the conductor and only penetrates up to the skin depth. Note that \$z\$ is in the direction perpendicular to the conductor. Usually what we really care about is the propagation in the direction parallel to the conductor.