Let's start with expanding the calculation of $E$:
$$
E(U,V)=\max_{|\psi\rangle}\sqrt{\langle\psi|(U-V)^\dagger(U-V)|\psi\rangle}.
$$
Clearly, we want $|\psi\rangle$ to be the eigenvector with maximum eigenvalue of
$$
2I-V^\dagger U-U^\dagger V.
$$
Let's note that
If $|\psi\rangle$ is an eigenvector of $U^\dagger V$, then the eigenvalue must be of the form $e^{i\theta}$ because $U^\dagger V$ is unitary
Furthermore, it is also an eigenvector of $V^\dagger U$, but with eigenvalue $e^{-i\theta}$ just by rearranging (multiplying by $V^\dagger U$) $U^\dagger V|\psi\rangle=e^{i\theta}|\psi\rangle$.
For a single-qubit rotation where $\text{det}(U)=\text{det}(V)=1$, there is a second eigenvector $|\psi^{\perp}\rangle$ whose eigenvalue is the conjugate (i.e. $e^{-i\theta}$ for $U^\dagger V$ and $e^{i\theta}$ for $V^\dagger U$) because the eigenvalues must have a product equal to 1.
So,
$$
E(U,V)=\sqrt{2-e^{i\theta}-e^{-i\theta}}=2\cos\frac{\theta}{2}
$$
Now let's consider $D(U,V)$.
$$
D(U,V)=\text{Tr}\left(\sqrt{(U-V)^\dagger(U-V)}\right).
$$
When taking the trace, we can use any basis we want. Let's use the eigenbasis of $2I-V^\dagger U-U^\dagger V$,
$$
D(U,V)=\langle\psi|\sqrt{(U-V)^\dagger(U-V)}|\psi\rangle+\langle\psi^\perp|\sqrt{(U-V)^\dagger(U-V)}|\psi^\perp\rangle.
$$
Square roots maintain the diagonalisation, which means we can do
$$
D(U,V)=\sqrt{\langle\psi|(U-V)^\dagger(U-V)|\psi\rangle}+\sqrt{\langle\psi^\perp|(U-V)^\dagger(U-V)|\psi^\perp\rangle},
$$
and we can now make use of the eigenvector relations
\begin{align*}
D(U,V)&=\sqrt{2+e^{i\theta}+e^{-i\theta}}+\sqrt{2+e^{-i\theta}+e^{i\theta}} \\
&=4\cos\frac{\theta}{2} \\
&=2E(U,V).
\end{align*}
For a completely different way to tackle the problem, let $U=e^{i\theta\underline{n}\cdot\underline{\sigma}}$ and $V=e^{i\phi\underline{m}\cdot\underline{\sigma}}$. Let's multiply out $U^\dagger V$. We get
$$
U^\dagger V=\cos\theta\cos\phi I-i\underline{n}\cdot\underline{\sigma}\cos\phi\sin\theta+i\underline{m}\cdot\underline{\sigma}\sin\phi\cos\theta+(\underline{n}\cdot\underline{\sigma})\cdot(\underline{m}\cdot\underline{\sigma})\sin\theta\sin\phi.
$$
Similarly,
$$
V^\dagger U=\cos\theta\cos\phi I+i\underline{n}\cdot\underline{\sigma}\cos\phi\sin\theta-i\underline{m}\cdot\underline{\sigma}\sin\phi\cos\theta+(\underline{m}\cdot\underline{\sigma})\cdot(\underline{n}\cdot\underline{\sigma})\sin\theta\sin\phi.
$$
So, when we add these two terms together, the middle two terms cancel immediately. The last term requires a little more thought. Note that
$$
(\underline{n}\cdot\underline{\sigma})\cdot(\underline{m}\cdot\underline{\sigma})=(\underline{n}\cdot\underline{m})I+(\underline{n}\times\underline{m})\cdot\underline{\sigma}.
$$
Also recal that the cross product is antisymmetric, so when we swap the order of terms, we get a negative sign. That means that the cross product term will also cancel. Hence
$$
U^\dagger V+V^\dagger U=2I(\cos\theta\cos\phi+\underline{n}\cdot\underline{m})\sin\theta\sin\phi).
$$
Overall, we have
$$
2I-U^\dagger V-V^\dagger=2I(1-\cos\theta\cos\phi-\underline{n}\cdot\underline{m}),
$$
which makes both the maximum eigenvector and the trace very easy to deal with and, critically, the trace of $I$ is double the maximum eigenvector.
