= Solution
We prove <Kantorovich duality theorem> here by a positive-functional extension argument; it produces the minimizing <transport plan> at the same time as equality of the values. Let $Y=X\times X$ and $c(x,y)=d(x,y)$. This is a bounded <continuous function> because $X$ is a compact <metric space>. Write $C(Y)$ for the real <space of continuous functions on a compact space>, and let
$$
S=\{s\in C(Y):s(x,y)=f(x)+g(y),\ f,g\in C(X)\}.
$$
On this <vector subspace> define $\ell(s)=\int f\,d\mathbf P+\int g\,d\mathbf Q$. It is well-defined: two representations differ by a constant in the first variable and the opposite constant in the second, and both measures have mass one. It is positive, since $s\geq0$ implies $\min f+\min g\geq0$ and hence $\ell(s)\geq0$. Also $\ell(1)=1$.
Define the majorant envelope, which is a <sublinear functional>:
$$
p(h)=\inf\{\ell(s):s\in S,\ s\geq h\}\qquad(h\in C(Y)).
$$
Constants majorize every $h$, and positivity gives $\min h\leq p(h)\leq\max h$, so this quantity is finite. Taking approximate minimizing majorants proves subadditivity; rescaling majorants proves positive homogeneity. Moreover $p(s)=\ell(s)$ for $s\in S$, and subtracting $s$ from a majorant proves
$$
p(h+s)=p(h)+\ell(s).
$$
The dual value in the question is exactly
$$
m_d=\sup\{\ell(s):s\in S,\ s\leq c\}=-p(-c).
$$
Subadditivity at $c+(-c)=0$ gives $m_d\leq p(c)$.
On $S+\mathbb Rc$, assign the value $\ell(s)+t m_d$ to $s+tc$. If $c\notin S$ the representation is unique. If $c\in S$, then $m_d=\ell(c)$, so the assignment is still well-defined. It is dominated by $p$: for $t\geq0$, use $t m_d\leq t p(c)=p(tc)$; for $t<0$, positive homogeneity gives $p(tc)=(-t)p(-c)=t m_d$. Together with the translation identity, these verify domination in both cases.
The real <Hahn-Banach theorem> extends this functional to $T:C(Y)\to\mathbb R$ with $T\leq p$. If $h\geq0$, then $p(-h)\leq\ell(0)=0$, so $T(h)\geq0$. Thus $T$ is a <positive linear functional>, with $T(1)=1$ and $T(c)=m_d$. Positivity also gives $|T(h)|\leq\|h\|_\infty$, so $T$ is continuous. The <Riesz-Markov-Kakutani representation theorem> supplies a Borel <probability measure> $\pi_0$ on the compact space $Y$, satisfying $T(h)=\int_Yh\,d\pi_0$.
For every $f\in C(X)$, its pullback $f(x)$ belongs to $S$, so $\int f(x)\,d\pi_0=\int f\,d\mathbf P$. Likewise $\int g(y)\,d\pi_0=\int g\,d\mathbf Q$. Uniqueness in the <Riesz-Markov-Kakutani representation theorem> shows that these are exactly the two <marginal distributions>. Finally, every feasible pair $f+g\leq c$ gives a lower bound for the cost of every <transport plan>, by integration. Our constructed plan achieves that bound:
$$
\boxed{\int_{X\times X}d(x,y)\,d\pi_0(x,y)
=m_d=\min_{\pi\text{ with marginals }\mathbf P,\mathbf Q}\int d\,d\pi.}
$$
This <Kantorovich duality by positive extension> establishes the requested existence and equality without assuming an optimal plan in advance.
The metric structure additionally gives the <Kantorovich–Rubinstein theorem> formulation. For a feasible pair define $h(x)=\inf_y\{d(x,y)-g(y)\}$. The triangle inequality for $d$ makes $h$ one-<Lipschitz continuous>, $h\geq f$, and $h(y)\leq-g(y)$. Hence the dual objective is at most $\int h\,d\mathbf P-\int h\,d\mathbf Q$. Conversely $(h,-h)$ is feasible for every one-Lipschitz $h$. Therefore
$$
m_d=\sup_{\operatorname{Lip}(h)\leq1}\left(\int h\,d\mathbf P-\int h\,d\mathbf Q\right).
$$
Normalizing $h(x_0)=0$ for a fixed $x_0$ gives a uniformly bounded equicontinuous family; it is closed and compact in the uniform topology by the <Arzela-Ascoli theorem>. The objective is uniformly continuous on this family, so the supremum is attained as well. In particular, $m_d$ is the first <Wasserstein distance> between the two measures.
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