Let $\mu, \nu, \omega \in \mathcal P_p (X)$. Let $\pi_1 \in \Pi (\mu, \nu)$ and $\pi_2 \in \Pi (\mu, \nu)$ be optimal, i.e.,
$$
W^p_p (\mu, \nu) := \int_{X \times Y} |x-y|^p \mathrm d \pi_1 (x, y) \quad \text{and} \quad W^p_p (\nu, \omega) := \int_{Y \times Z} |y-z|^p \mathrm d \pi_2 (y, z).
$$
Let $P^{X \times Y}$ and $P^{Y \times Z}$ be the projection maps from $X \times Y \times Z$ to $X \times Y$ and $Y \times Z$ respectively. Let $\gamma \in \mathcal P(X \times Y \times Z)$ such that $P^{X \times Y}_\sharp \gamma = \pi_1$ and $P^{Y \times Z}_\sharp \gamma = \pi_2$. Here $P^{X \times Y}_\sharp \gamma$ is the push-forward of $\gamma$ by $P^{X \times Y}$. Such $\gamma$ does exists by gluing lemma. Let $\pi = P^{X \times Z}_\sharp \gamma$. It's easy to prove that $\pi \in \Pi(\mu, \omega)$. Finally,
$$
\begin{align}
W_p(\mu, \omega) &\le \left [ \int_{X \times Z} |x-z|^p \mathrm d \pi (x, z) \right ]^{1/p} \\
&= \left [ \int_{X \times Y \times Z} |(x-y) - (y-z)|^p \mathrm d \gamma(x,y,z) \right ]^{1/p} \\
&\le \left [ \int_{X \times Y} |x-y|^p \mathrm d \pi_1 (x, y) \right ]^{1/p} + \left [ \int_{Y \times Z} |y-z|^p \mathrm d \pi_2 (y, z) \right ]^{1/p} \\
&= W_p(\mu, \nu) + W_p(\nu, \omega).
\end{align}
$$
This completes the proof.
Best Answer
Let $(\mu_n)_{n\ge1}$ be a sequence in $\Delta(C)$. It is obviously tight because $\mu_n(K)=1\ge1-\varepsilon$ for the compact $K:=C$ and for all $n\ge1$ and $\varepsilon>0$. By Prokhorov's theorem, we know that $(\mu_n)_{n\ge1}$ has some limit point $\mu$ w.r.t. the topology of weak convergence. By Skorokhod's representation theorem, there exist a probabiltiy space $(\Omega,\mathcal A,\mathbb P)$ and random variables $X,X_1,X_2,\ldots:\Omega\to C$ such that $X$ has law $\mu$, $X_n$ has law $\mu_n$ for every $n\ge1$ and $$d(X_n,X)\xrightarrow[n\to\infty]{}0\quad\text{$\mathbb P$-a.s.}$$ Now because $C$ is compact, there exists a constant $a>0$ such that $d(X_n,X)\le a$, thus $d(X_n,X)^p\le a^p$ for every $n\ge1$. Hence $$\mathbb E[d(X_n,X)^p]\xrightarrow[n\to\infty]{}0$$ by dominated convergence. Thus we have couplings $\gamma_n\in\Pi(\mu_n,\mu)$ such that $$\left(\int d(x,y) ^p\,\mathrm d\gamma_n(x,y)\right)^{\!\frac1p}\xrightarrow[n\to\infty]{}\:0.$$ Since $W_p(\mu_n,\mu)$ is the infimum of the left-hand side over $\Pi(\mu_n,\mu)$, we deduce that $$W_p(\mu_n,\mu)\xrightarrow[n\to\infty]{}0,$$ which means that $\mu$ is also a limit point of $(\mu_n)_{n\ge1}$ for the Wasserstein-$p$ metric. Hence $(\Delta(C),W_p)$ is compact.