Before presenting my question (which I already formulate in the title of this post) is important to establish the context of my problem:
Definition: The $p$-Wasserstein metric $W_{p}(\mu,\nu)$ between $\mu,\nu\in\mathcal{P}_{p}(\Xi)$ is defined by $$W_{p}^{p}(\mu,\nu):=\min_{\Pi\in\mathcal{P}(\Xi\times\Xi)}\left\{\int_{\Xi\times\Xi}d^{p}(\xi,\zeta)\Pi(d\xi,d\zeta)\: :\: \Pi(\cdot \times\Xi)=\mu(\cdot),\: \Pi(\Xi\times\cdot)=\nu(\cdot)\right\}$$ where $$\mathcal{P}_{p}(\Xi):=\left\{\mu\in\mathcal{P}(\Xi)\: :\: \int_{\Xi}d^{p}(\xi,\zeta_{0})\mu(d\xi) < \infty\ \mbox{for some }\zeta_{0}\in\Xi\right\}$$ where $d$ is a metric on $\Xi$.
The $p$-Wasserstein metric (with $p\geq 1$) is also defined for any measure in $\mathcal{P}(\Xi)$ the space of all the measures of probability, the only difference is that in that set it can take values as infinite.
Given $\varepsilon>0$ and $\mu\in\mathcal{P}_{p}(\Xi)$ we define $$\mathcal{B}_{\varepsilon}^{p}(\mu):=\left\{\nu\in\mathcal{P}_{p}(\Xi)\:|\: W_{p}(\mu,\nu)\leq \varepsilon \right\}$$ and $$\mathcal{B}_{\varepsilon}(\mu):=\left\{\nu\in\mathcal{P}(\Xi)\:|\: W_{p}(\mu,\nu)\leq \varepsilon \right\}.$$
The question: I want to know if the following is correct: $$\mathcal{B}_{\varepsilon}(\mu)=\mathcal{B}_{\varepsilon}^{p}(\mu).$$
My attempt: We know that $\mathcal{P}_{p}(\Xi)\subset \mathcal{P}(\Xi)$, then $$\mathcal{B}_{\varepsilon}^{p}(\mu)\subseteq\mathcal{B}_{\varepsilon}(\mu).$$ For other hand, if $\nu\in \mathcal{B}_{\varepsilon}(\mu)$ then $W_{p}(\nu,\mu)\leq \varepsilon$ then by infimum definition there exists $\Pi\in\mathcal{P}(\Xi\times\Xi)$ such that $\Pi(\cdot\times \Xi)=\nu$, $\Pi(\Xi\times \cdot)=\mu$ and $$\int_{\Xi\times\Xi}d^{p}(\xi,\zeta)\Pi(d\xi,d\zeta)\leq \varepsilon.\tag{*} $$ Therefore, due to the convexity of $x^{p}$, $p$ is positive integer with $p\geq 1$, $\mu\in\mathcal{P}_{p}(\Xi)$ and the triangle inequality we have $$ \begin{array}{rcl} \int_{\Xi}d^p(\xi_{0},\xi)\nu(d\xi) &=& \int_{\Xi\times\Xi} d^p(\xi,\xi_{0})\Pi(d\xi,d\zeta) \\ &=& 2^{p} \int_{\Xi\times\Xi} \left(\frac{d(\xi,\xi_{0})}{2}\right)^{p}\Pi(d\xi,d\zeta) \\ &\leq & 2^{p-1}\int_{\Xi\times\Xi} d^p(\xi_{0},\zeta)\Pi(d\xi,d\zeta)+ 2^{p-1}\int_{\Xi\times\Xi} d^p(\xi,\zeta)\Pi(d\xi,d\zeta) \\ &\leq& 2^{p-1}\int_{\Xi\times\Xi} d^p(\xi_{0},\zeta)\Pi(d\xi,d\zeta)+ 2^{p-1}\varepsilon \\ &=& 2^{p-1}\int_{\Xi} d^p(\xi_{0},\zeta)\mu(d\zeta)+ 2^{p-1}\varepsilon \\ &<&\infty. \end{array} $$ Therefore, $\nu\in\mathcal{P}_{p}(\Xi)$, so $\nu\in\mathcal{B}_{\varepsilon}^{p}(\mu)$.
Remark: The previous attempt was made after several attempts, when I was very tired, so it is highly likely that I can go wrong. Is my attempt correct? Do I have an error in my attempt? If my attempt is incorrect, I would appreciate it if you would give me an answer to my question.
Addemdum:
I calmly set myself the task of analyzing my proof, I realized that it has an error, although the fact that I try to show is true. Questions like the one I proposed are very common in the knowledge exams of Masters or Phd's, if you want the correct demonstration of the fact that I propose let me know, but ideally you should identify the error of my attempt, it is a good exercise.