Upper bound total variation by Wasserstein distance for continuous distance

No. One should realize that the transportation and the total variation distances metrize two quite different topologies. Even if the measures are equivalent (i.e., absolutely continuous with respect to each other), one can still easily have examples when the transportation distance is arbitrarily close to 0, whereas the total variation distance is arbitrarily close to 2. Bounding the Radon-Nikodym derivatives is not of much help either as, once again, one can easily have examples of measures arbitrarily close in the transportation metric, whereas their total variation distance will be the one dictated by the density bounds. However, if you talk about measures from a certain specific class, then the situation may be different.

If both probability measures have smooth densities, the total variation can be bounded by Wasserstein (even by a weaker metric Levy-Prokhorov), see this paper.

More precisely, see Lemma 5.1 and proof of Theorem 2.1. The basic idea is given below.

If both $p$ and $q$ are smooth densities, $d_V(p, p_\gamma)$ and $d_V(q, q_\gamma)$ are sufficiently small for every small gamma, where $d_V$ is the total variation and $p_\gamma$ is a convolution of p and the uniform density as defined in the paper.

Note that $d_V(p, q) < d_V(p, p_\gamma) + d_V(p_\gamma, q_\gamma) + d_v(q_\gamma, q)$ by the triangle inequality.

Lemma 5.1 guarantees that if $p$ and $q$ are close in Levy-Prokhorov metric, then $d_V(p_\gamma, q_\gamma)$ is also small provided that $\gamma$ is much bigger than the Levy-Prochorov distance.