Budget constraint hyperplanes
The set $\{\mathbb{p}:\exists \alpha \in (0,1), \mathbb{p}_1=\alpha b_i, \mathbb{p}_2=(1-\alpha)b_i\}$ coincides with the hyperplane $\{\mathbb{p}:p_1+p_2=b_i\}$, therefore you are not finding extra hyperplanes by doing this and varying $\alpha$.

More generally, if you fix $x=(x_1,x_2)$ and $b_i$, you obtain one hyperplane characterized by the equation $\{\mathbb{p}:p_1 x_1+p_2 x_2=b_i\}$. Since there are finitely many such $x$ (and $b_i$ is given), there are finitely many such hyperplanes.