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Edit: (This question may be better suited for math.stackexchange.)

This is true. It suffices to show the map $\mu\rightarrow W_{1}(\mu,\nu)$ is convex. Let $\mu_{i}\in\mathcal{P}(\Xi)$ and $\psi^{*}_i\in\Gamma(\mu_{i},\nu)$ be optimal transport plans between $\mu_{i}$ and $\nu$ for each $i=1,2$. Then $\lambda\psi_{1}^{*}+(1-\lambda)\psi_{2}^{*}\in\Gamma(\lambda\mu_{1}+(1-\lambda)\mu_{2},\nu)$ and so

$W_{1}(\lambda\mu_{1}+(1-\lambda)\mu_{2},\nu)=\inf\limits_{\gamma\in\Gamma(\lambda\mu_{1}+(1-\lambda)\mu_{2},\nu)} \int d(x,y) d\gamma(x,y)\leq \\ \int d(x,y) d(\lambda\psi_{1}^{*}+(1-\lambda)\psi_{2}^{*})(x,y) = \lambda W_{1}(\mu_{1},\nu)+(1-\lambda)W_{1}(\mu_{2},\nu)$

which gives convexity. Thus, $$W_{1}(\lambda\mu_{1}+(1-\lambda)\mu_{2},\nu)\leq \lambda W_{1}(\mu_{1},\nu)+(1-\lambda)W_{1}(\mu_{2},\nu)\leq \theta.$$

When restrictions are made such as replacing $\mathcal{P}(\Xi)$ with some other set of probability measures, as is done in the paper here, the answer can be no:

[For $\mathcal{P}(\Xi)$ replaced with $\mathcal{N}(\Xi)$, Gaussian distributions]...the Wasserstein ambiguity set $P$ is nonconvex (as mixtures of normal distributions are generically not normal).

Edit: (This question may be better suited for math.stackexchange.)

This is true. It suffices to show the map $\mu\rightarrow W_{1}(\mu,\nu)$ is convex. Let $\mu_{i}\in\mathcal{P}(\Xi)$ and $\psi^{*}_i\in\Gamma(\mu_{i},\nu)$ be optimal transport plans between $\mu_{i}$ and $\nu$ for each $i=1,2$. Then $\lambda\psi_{1}^{*}+(1-\lambda)\psi_{2}^{*}\in\Gamma(\lambda\mu_{1}+(1-\lambda)\mu_{2},\nu)$ and so

$W_{1}(\lambda\mu_{1}+(1-\lambda)\mu_{2},\nu)=\inf\limits_{\gamma\in\Gamma(\lambda\mu_{1}+(1-\lambda)\mu_{2},\nu)} \int d(x,y) d\gamma(x,y)\leq \\ \int d(x,y) d(\lambda\psi_{1}^{*}+(1-\lambda)\psi_{2}^{*})(x,y) = \lambda W_{1}(\mu_{1},\nu)+(1-\lambda)W_{1}(\mu_{2},\nu)$

which gives convexity. Thus, $$W_{1}(\lambda\mu_{1}+(1-\lambda)\mu_{2},\nu)\leq \lambda W_{1}(\mu_{1},\nu)+(1-\lambda)W_{1}(\mu_{2},\nu)\leq \theta.$$

When restrictions are made such as replacing $\mathcal{P}(\Xi)$ with some other set of probability measures, as is done in the paper here, the answer can be no:

[For $\mathcal{P}(\Xi)$ replaced with $\mathcal{N}(\Xi)$, Gaussian distributions]...the Wasserstein ambiguity set $P$ is nonconvex (as mixtures of normal distributions are generically not normal).

Edit: (This question may be better suited for math.stackexchange.)

This is true. It suffices to show the map $\mu\rightarrow W_{1}(\mu,\nu)$ is convex. Let $\mu_{i}\in\mathcal{P}(\Xi)$ and $\psi^{*}_i\in\Gamma(\mu_{i},\nu)$ be optimal transport plans between $\mu_{i}$ and $\nu$ for each $i=1,2$. Then $\lambda\psi_{1}^{*}+(1-\lambda)\psi_{2}^{*}\in\Gamma(\lambda\mu_{1}+(1-\lambda)\mu_{2},\nu)$ and so

$W_{1}(\lambda\mu_{1}+(1-\lambda)\mu_{2})=\inf\limits_{\gamma\in\Gamma(\lambda\mu_{1}+(1-\lambda)\mu_{2},\nu)} \int d(x,y) d\gamma(x,y)\leq \\ \int d(x,y) d(\lambda\psi_{1}^{*}+(1-\lambda)\psi_{2}^{*})(x,y) = \lambda W_{1}(\mu_{1},\nu)+(1-\lambda)W_{1}(\mu_{2},\nu)$

which gives convexity. Thus, $$W_{1}(\lambda\mu_{1}+(1-\lambda)\mu_{2},\nu)\leq \lambda W_{1}(\mu_{1},\nu)+(1-\lambda)W_{1}(\mu_{2},\nu)\leq \theta.$$

When restrictions are made such as replacing $\mathcal{P}(\Xi)$ with some other set of probability measures, as is done in the paper here, the answer can be no:

[For $\mathcal{P}(\Xi)$ replaced with $\mathcal{N}(\Xi)$, Gaussian distributions]...the Wasserstein ambiguity set $P$ is nonconvex (as mixtures of normal distributions are generically not normal).

Edit: (This question may be better suited for math.stackexchange.)

This is true. It suffices to show the map $\mu\rightarrow W_{1}(\mu,\nu)$ is convex. Let $\mu_{i}\in\mathcal{P}(\Xi)$ and $\psi^{*}_i\in\Gamma(\mu_{i},\nu)$ be optimal transport plans between $\mu_{i}$ and $\nu$ for each $i=1,2$. Then $\lambda\psi_{1}^{*}+(1-\lambda)\psi_{2}^{*}\in\Gamma(\lambda\mu_{1}+(1-\lambda)\mu_{2},\nu)$ and so

$W_{1}(\lambda\mu_{1}+(1-\lambda)\mu_{2},\nu)=\inf\limits_{\gamma\in\Gamma(\lambda\mu_{1}+(1-\lambda)\mu_{2},\nu)} \int d(x,y) d\gamma(x,y)\leq \\ \int d(x,y) d(\lambda\psi_{1}^{*}+(1-\lambda)\psi_{2}^{*})(x,y) = \lambda W_{1}(\mu_{1},\nu)+(1-\lambda)W_{1}(\mu_{2},\nu)$

which gives convexity. Thus, $$W_{1}(\lambda\mu_{1}+(1-\lambda)\mu_{2},\nu)\leq \lambda W_{1}(\mu_{1},\nu)+(1-\lambda)W_{1}(\mu_{2},\nu)\leq \theta.$$

When restrictions are made such as replacing $\mathcal{P}(\Xi)$ with some other set of probability measures, as is done in the paper here, the answer can be no:

[For $\mathcal{P}(\Xi)$ replaced with $\mathcal{N}(\Xi)$, Gaussian distributions]...the Wasserstein ambiguity set $P$ is nonconvex (as mixtures of normal distributions are generically not normal).

Edit: This question may be better suited for math.stackexchange. This is true, and also holds for all order $p$ mixed Wasserstein distances. For that case, see pg. 21 of the talk here by Corina Birghila, 'the mixed Wasserstein distance is convex in each component'. Thus, $$W_{1}(\lambda\mu_{1}+(1-\lambda)\mu_{2},\nu)\leq \lambda W_{1}(\mu_{1},\nu)+(1-\lambda)W_{1}(\mu_{2},\nu)\leq \theta.$$

When restrictions are made such as replacing $\mathcal{P}(\Xi)$ with some other set of probability measures, as is done in the paper here, the answer can be no:

[For $\mathcal{P}(\Xi)$ replaced with $\mathcal{N}(\Xi)$, Gaussian distributions]...the Wasserstein ambiguity set $P$ is nonconvex (as mixtures of normal distributions are generically not normal).

Edit: (This question may be better suited for math.stackexchange.)

This is true. It suffices to show the map $\mu\rightarrow W_{1}(\mu,\nu)$ is convex. Let $\mu_{i}\in\mathcal{P}(\Xi)$ and $\psi^{*}_i\in\Gamma(\mu_{i},\nu)$ be optimal transport plans between $\mu_{i}$ and $\nu$ for each $i=1,2$. Then $\lambda\psi_{1}^{*}+(1-\lambda)\psi_{2}^{*}\in\Gamma(\lambda\mu_{1}+(1-\lambda)\mu_{2},\nu)$ and so

$W_{1}(\lambda\mu_{1}+(1-\lambda)\mu_{2})=\inf\limits_{\gamma\in\Gamma(\lambda\mu_{1}+(1-\lambda)\mu_{2},\nu)} \int d(x,y) d\gamma(x,y)\leq \\ \int d(x,y) d(\lambda\psi_{1}^{*}+(1-\lambda)\psi_{2}^{*})(x,y) = \lambda W_{1}(\mu_{1},\nu)+(1-\lambda)W_{1}(\mu_{2},\nu)$

which gives convexity. Thus, $$W_{1}(\lambda\mu_{1}+(1-\lambda)\mu_{2},\nu)\leq \lambda W_{1}(\mu_{1},\nu)+(1-\lambda)W_{1}(\mu_{2},\nu)\leq \theta.$$

When restrictions are made such as replacing $\mathcal{P}(\Xi)$ with some other set of probability measures, as is done in the paper here, the answer can be no:

[For $\mathcal{P}(\Xi)$ replaced with $\mathcal{N}(\Xi)$, Gaussian distributions]...the Wasserstein ambiguity set $P$ is nonconvex (as mixtures of normal distributions are generically not normal).