Given two random variables X,Y with measures P,Q. Show that if $P(E) \le Q(E^\alpha) + \beta$ for all measurable $E\subset\mathbb{R}$ then $\mathbb{P}(d(X,Y)>\alpha)<\beta$.

Only hints please.

Attempt

For all $\varepsilon>0$ Strassen's theorem gives us measure $\mu$ on $\mathbb{R}^{2}$ s.t. $\mu(d(X,Y) >\alpha +\varepsilon)\leq \varepsilon+\beta$ and marginals P,Q. Now I want to relate the measures $\mathbb{P}$ and $\mu$.

We have $$\mathbb{P}(X\in A, Y\in B)-\mu(A\times B)=\mathbb{P}(X\in A)+\mathbb{P}( Y\in B)-\mathbb{P}(X\in A\cup Y\in B)-[\mu(A\times S)+\mu(S\times B)-\mu(A\times S\cup S\times B)]=\mu(A\times S\cup S\times B)-\mathbb{P}(X\in A\cup Y\in B)$$

Also,

$$\mu(d(x,y) < \alpha )=\mu(\bigcup_{x_{i}\in D} |x-x_{i}|< \alpha \times |y-x_{i}|< \alpha),$$

where D is the separable subset of X. I am trying to connect this with $\mathbb{P}(\bigcup_{x_{i}\in D} |X-x_{i}|< \alpha \cap |Y-x_{i}|< \alpha)$.