(a) The idea (similar to exercise 15): any open set is the product such that only finitely many open sets are not equal spaces. We can always separate the indexes of these spaces by some rational points. So let’s take all finite subsets of rationals in $[0,1]$ containing 0 and 1 (countably many), each such a subset generates a finite number of half-open intervals of the form [), and then take products of rational points such that they are the same across each interval (countably many).(b) Define $f$ as in the hint for a fixed interval. Note that all $f(\alpha)$ are different (indeed, for $\alpha\neq\beta$ : $\pi_\alpha^{-1}((a,b))\cap\pi_\beta^{-1}((b,+\infty))$ must have a dense point but lies within $\pi_\alpha^{-1}((a,b))$ only). Therefore, $f$ is an injection and $|J|\le|2^D|$ .