 Solution

The quotient $G\to D_{2n}$ adds the relation $r^n=1$. All four one-dimensional characters factor through some such quotient: choose even $n$ when $r$ acts as $-1$. A two-dimensional representation from part (b) factors through $D_{2n}$ exactly when ${\lambda }^n=1$. Hence the irreducible representations that never arise from a finite dihedral quotient are precisely the two-dimensional ones for which $\lambda$ is not a root of unity.