Why is the residual proton resonance in deuterated benzene a singlet?
To a very good approximation, $J_\ce{HD}$ in a deuterated molecule is related to $J_\ce{HH}$ in the corresponding protonated molecule by the gyromagnetic ratios:

$$\frac{J_\ce{HD}}{J_\ce{HH}} = \frac{\gamma_\ce{D}}{\gamma_\ce{H}} = \frac{\pu{41.066 * 10^6 rad s-1 T-1}}{\pu{267.522 * 10^6 rad s-1 T-1}} = 0.15$$

In benzene where $^3J_\ce{HH} = \pu{7.6 Hz}$ and $^4J_\ce{HH} = \pu{1.4 Hz}$ (Reich NMR tables), we would thus have $^3J_\ce{HD} \approx \pu{1.2 Hz}$ and $^4J_\ce{HD} \approx \pu{0.2 Hz}$. The four-bond meta coupling would be all but invisible, and the three-bond ortho coupling if resolved would give rise to a $1:2:3:2:1$ quintet.

On a lower-field spectrometer this relatively small coupling is usually not properly resolved. A suitable window multiplication, though, will reveal all. I don't have a d6-benzene spectrum on hand, but I do have a $\pu{500 MHz}$ d8-toluene spectrum here. The two spectra on the left are the aromatic residual peaks, but at the bottom-left a Gaussian window function has been used to improve resolution (at the cost of signal:noise ratio). This allows us to resolve the three-bond ortho couplings of $\pu{1.1 Hz}$.



At the bottom-right is the signal corresponding to residual $\ce{C6D5CD2H}$. Since the two-bond H–D coupling has a larger magnitude ($^2J_\ce{HD} = \pu{2.2 Hz}$), it is much easier to resolve, and can be seen even with the normal exponential window function (which sacrifices resolution for signal:noise ratio improvement).

Similarly, in d6-DMSO (shown below at $\pu{500 MHz}$), the two-bond coupling is more easily resolved.