Why should LTB dust be comoving? In many research papers about inhomogeneous cosmology, one often considers spherically symmetric (LTB) spacetimes where in the co-ordinate frame $(t,r,\theta,\varphi)$ wherein the metric assumes the form
$$ds^2=-dt^2+\frac{(Y'(r,t)dr)^2}{1-k(r)}+Y^2(r,t)(d\theta^2+\sin^2\theta d\varphi^2)\qquad (1) $$
one solves for the Einstein equations sourced by a comoving dust:
\begin{equation}T_{\mu\nu}(r,t) = \rho(r,t)dt^2 \qquad (2)\end{equation}
How is that motivated? 
(The objection against this choice is that in a metric in the above diagonal form (1) it does not seem that a dust $T_{\mu\nu}=\rho U_{\mu}U_\nu$, where $U$ has a non-vanishing radial component, can be brought in the form (2) by a $(r,t) \mapsto (r',t')$-diffeomorphism without destroying the form (1) of the metric) 
 A: I may be misunderstanding the question, but: --
"Comoving" is normally a description of coordinates or an observer, not a matter field. For example, the FLRW coordinates are considered comoving because an observer at constant $(r,\theta,\phi)$ is at rest with respect to the local matter. This property of the coordinates is not preserved under a diffeomorphism. That's why some coordinates are comoving and some are not. So here, the dust isn't comoving, the coordinates are comoving because they're moving with the dust.
A perfect fluid can be defined as a matter field that is completely characterized by its pressure and density, so that a frame exists in which the stress-energy tensor has the form $\operatorname{diag}(\rho,P,P,P)$. A dust is one in which the pressure vanishes. So since the matter content of the LTB spacetime is dust, it is guaranteed that at every point in spacetime, there is a frame in which the stress-energy has the form (2) that you give. This is the comoving frame.

(The objection against this choice is that in a metric in the above diagonal form (1) it does not seem that a dust $T_{\mu\nu}=\rho U_{\mu}U_\nu$, where $U$ has a non-vanishing radial component, can be brought in the form (2) by a $(r,t) \mapsto (r',t')$-diffeomorphism without destroying the form (1) of the metric)

I'm not sure if I'm correctly understanding your point, but I think what you're saying just amounts to a statement that it's nontrivial to find solutions to the Einstein field equations. If you take an LTB solution and then alter it by changing the state of motion of the dust, it won't be a solution anymore, for the same form of the metric.
