Your first argument is completely wrong. The second one is partially correct. Similar to benzene, pyridine is aromatic but the lone pair of $\ce{N}$ is not involved in the $(4n+2)$-$\pi$ electron system. Instead, the lone pair is in a $\mathrm{sp^2}$ hybrid orbital in the same plane away from conjugated $(4n+2)$-$\pi$ electron system. As a consequence, the $\ce{N}$ atom is weakly basic ($\mathrm{p}K_\mathrm{a} = 5.2$), because this lone pair of $\ce{N}$ can function as a source of electrons (in the Lewis base) and is in a $\sigma$-orbital. Therefore it is not a part of the aromatic $\pi$-system. Only the $\ce{N}$ atom is part of $\pi$-systemas it makes $\ce{C=N}$ bond with the system. Once protonated, it is still in the $\pi$-system and the pyridine nucleus maintains its cyclic, planar, conjugated, $6\pi$-electron system.
The lone pair of Pyrrole, on the other hand, is part of conjugated $(4n+2)$-$\pi$ electron system. Once protonated, pyrrole loses its aromaticity due to the loss of lone pair to proton ($\ce{H+}$). Therefore, this would result in protonation being unfavourable, and protonated pyrrole becomes unstable. As a result, pyrrole is a much weaker base than pyridine ($\mathrm{p}K_\mathrm{a} = -3.8$).
Pyrrolidine is technically tetrahydropyrrole. Unlike pyridine and pyrrole, the nitrogen atom of pyrrolidine is $\mathrm{sp^3}$ hybridized. Thus, it has more $\mathrm{p}$ character while $\ce{N}$ of pyridine and pyrrole have more $\mathrm{s}$ character comparatively. As a result, pyridine and pyrrole hold their lone pair electrons tighter than that of pyrrolidine. Consequently, Pyrrolidine has its lone pair readily available to incoming protons compared to that of pyridine and pyrrole. Therefore, pyrrolidine is the strongest base ($\mathrm{p}K_\mathrm{a} = 11.3$) among three.
Source of $\mathrm{p}K_\mathrm{a}$ values
