= Solution
Given any admissible <transport map> $T$, form its graph <transport plan>
$$
\pi_T=(\operatorname{Id},T)_\#\mu.
$$
For <Borel sets> $A\subseteq X$ and $B\subseteq Y$, the definition of a <pushforward measure> gives
$$
\pi_T(A\times Y)=\mu(A),\qquad
\pi_T(X\times B)=\mu(T^{-1}(B))=\nu(B).
$$
Hence $\pi_T\in\Pi(\mu,\nu)$. Integration against a <pushforward measure> also gives
$$
\mathbb K(\pi_T)=\int_X c(x,T(x))\,d\mu(x)=\mathbb M(T).
$$
The <Kantorovich optimal transport problem> therefore has at least all the competitors of the <Monge optimal transport problem>, with exactly the same costs. Consequently
$$
\boxed{\inf_{T_\#\mu=\nu}\mathbb M(T)\ \geq\ \inf_{\pi\in\Pi(\mu,\nu)}\mathbb K(\pi).}
$$
If there is no admissible <transport map>, the left side is $+\infty$ by the convention $\inf\varnothing=+\infty$, so the conclusion still holds. No existence of an optimizer is needed.
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