I would like to show that for the curves $[0,1]\ni t\mapsto \mu_t:=(1-t)\delta_x+t\delta_y$ we have $$W_2(\mu_t,\mu_s)=\sqrt{|t-s|}d(x,y)$$ where $W(\cdot,\cdot)$ is the $2$-Wasserstein distance.
My attempt : Using the convexity of $W_2^2$ in both variables and the fact that $W(\delta_x,\delta_y)=d(x,y)$ and $d(x,x)=d(y,y)=0$ we get that $$W_2^2(\mu_t,\mu_s)=W_2^2((1-t)\delta_x+t\delta_y,(1-s)\delta_x+s\delta_y)\le (1-t)sW_2^2(\delta_x,\delta_y)+t(1-s)W_2^2(\delta_y,\delta_x)=((1-t)s+t(1-s))d(y,x)$$ but I cannot conclude.