A general note is that the answer depends heavily on the properties of $\mu$. 

First a note that in general $d(A,B) \not \le W_p(\mu|_A,\mu|_B)$ for $p\in[1,\infty)$. Though it's true for the case $p = \infty$. Here the example: Let $\mu_\lambda = (1-\lambda) \delta_x + \lambda \delta_y$ for distinct $x,y \in X$. Choose $A=\{x\}$ and $B=\{x,y\}$. Then $\mu_\lambda|_A$ is $\delta_x$ and $\mu_\lambda|_B$ is $\mu_\lambda$. Thus $d(A,B) = d(x,y)$ and $W_p(\mu|_A,\mu|_B) = \lambda^{\frac{1}{p}}d(x,y)$.

The case $p=\infty$, i.e. $W_\infty(\mu,\nu) = \inf \|d\|_{L^\infty(\pi)} $, it is true. 

But one there is no lower bound for the Hausdorff distance $d(\cdot,\cdot)$ w.r.t. to any Wasserstein distance $W_p$ if there is at least one non-isolated point (as $W_p \le W_\infty$ it suffices to get counterexamples for $p=\infty$).  

Choose $x_n \to x$ and $\mu = \frac{1}{2}\delta_x + \sum_{n\in \mathbb{B}} \lambda_n \delta_{x_n}$ for some $\lambda_n \ge 0$ with $\sum_{n\in \mathbb{B}} \lambda_n = \frac{1}{2}$. Assume, in addition, $\lambda_1 > 2 \lambda_n$ for $n>1$. Now Choose $A = \{x, x_1\}$ and $B_n = \{x_n,x_1\}$. Then $d(A,B_n) = d(x,x_n) \to 0$. However, $W_\infty(\mu|_A, \mu|_{B_n}) \ge \inf \{d(x,x_1),d(x_n,x_1)\}$ because $\mu|_A(\{x\}) > \frac{1}{2} > \mu|_{B_n}(\{x_n\})$.