This doesn't seem quite correct. For one thing the final condition does not make sense as written. It should probably be  " It is not the case that $d_{2i}=a_{2i+1}$ (so $d_{2i-1}=0$) for all large enough $i$ " (equivalently "$d_{2i} \lt a_{2i+1}$ infinitely often") .  One would also want the similar condition for the situation $d_{2i-1}=a_{2i}.$ 

Perhaps $d_0$ can be any integer, positive or negative. Otherwise there is certainly an upper bound (perhap$(a_0+1)\theta_1$) on the  $\beta$  which can be expressed. 

HOWEVER, something along these lines should be true. You could likely discover then prove it by (imagining that) you are looking at an example such as $\sqrt{3}=[3;3,3,3,\cdots]$, considering the sums with $d_i=0$ for $i \gt N$ and where they are: The following would seem to  need to be true along with the obvious extensions. Only the indicated inequalities actually need to be justified.  $$0 \lt \theta_0 \lt 2\theta_0 \lt 3\theta_0$$  $$(d_0-1)\theta_0 \lt^? d_0\theta_0+3\theta_1 \lt d_0\theta_0+2\theta_1 \lt d_0\theta_0+\theta_1 \lt d_0\theta_0 $$ 

$$d_1\theta_1 \lt d_1\theta_1+\theta_2 \lt d_1\theta_1+2\theta_2\lt d_1\theta_1+3\theta_2  \lt^? (d_1-1)\theta_1$$

Describe what happens (for general $\alpha$), prove it using the properties of continued fractions, figure out what breaks down with $d_i=a_{i+1}$ and $d_{i-1} \ne 0$ and with having the (correct) final conditions fail.