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@DmitriPanov, you are right, I missed that point. A quick check, only gave that the image of closed ball is the closed ball, i.e. $f(\bar U) = \overline{f(U)}$. Via a contradiction argument it is also possible to prove that a half space is mapped to a half space and that $f$ has to be surjective. This would rescue the convexity property. Currently I see a problem that the two touching points at the boundary of the big ball might not get mapped onto the boundary of the image of the large ball. Maybe the half-space property is sufficient to prove this. I'll try to fix it as soon as possible.
If $f|_S$ is uniformly continuous then it can be natuarlly extended to a uniformly continuous function on the closure of $S$. Hence $f$ itself has to be uniformly continuous. If you weaken the assumption to "there is a sequence of sets $S_n$ such that their union is dense in $R$ and f restricted to $S_n$ is uniformly continuous on $S_n$" then it's true for measurable functions by Lusin's Theorem.
Note that the quotient space (spaces of equivalence classes induced by x~y iff d(x,y) is a metric space with metric induced by d. However it might be a very bad metric space, like one not satisfying Prokhorov.
Sorry by “this setting” I meant Polish spaces. There are metric spaces that are not Polish (any non-separable Hilbert/Banach space). There are even metric spaces that are incomplete but Polish (think of the sphere and remove a point, this is homeomophic to Euclidean space hence incomplete with the spherical metric but complete with the Euclidean one).
In this setting you may even assume the pseudometric is only lower semicontinuous as it is exactly the setting of Villani’s book and the majority of papers on optimal transport. A dual solution exists as well, but there might not be a dual solution if the cost function is infinite somewhere.
In my opinion adding a delta might only lead to a value that is equal to $p \cdot b(\mathcal{N},\nu)$ where $b(\mathcal{N},\nu) = \inf \int d^2(y_0,y)d\nu(y)$.
Here some idea of a general construction: There is no relationship of $T$ and $\nu$ meaning continuous map would do the job. Via approximation it's enough to look an open domain $V$ of full $\nu$-measure that is starshaped from a point fixed point. Now construct a smooth map $T_n$ that maps an (starshaped) closed domain $K_n$ of $\mathcal{M}$ of measure $\mu(\mathcal{M})-\frac{1}{n}$ into the chosen domain and is constant outside of $K_n$. Altering the map inside of $K_n$ a bit one can assume the push-forward of the smooth part of $\mu$ is equal to (the smooth part) of $\nu$ on $T(K_n)$.
Your problem does not demand continuous transport of $\mu$ to $\nu$. Any coupling and any continuous/smooth $T$ would do. Hence let $T(r,s)=s$ where $D^1$ is the unit disk and $S^1$ is parametrized by $[-1,1)$. The push-forward is a absolutely continuous measure on the circle with density having a bump at $0$ and being zero at $\pm 1$. It's even possible to find a continuous map such that the push forward is exactly the uniform measure on $S^1$. Choose a coupling concentrated on $\{ (r,s,s) | r,s \in [-1,1]\}$. Problems arise if $\dim \mathcal{M} < \dim \mathcal{N}$.
