You are right that that particular base $B$ is uncountable; your mistake is thinking that because $B$ is uncountable, every base for the usual topology on $\Bbb R^n$ must be uncountable. If you keep only those members of $B$ for which the endpoints $\alpha_i$ and $\beta_i$ are all rational, you’ll have a countable family that is still a base for the topology.
You are right that $C_i\neq \bigcup X$ and that rather $C_i \bigcup {X}$. I think it is easiest to prove $C_i$ is countable by induction on $i$. For $i=0$ this is trivial (in fact $C_0$ has only one element: the empty set), and it's still pretty trivial for $i=1$ (since $C_1$ is more or less $\mathbb{Z}$ itself). Now, supposing $C_n$ is countable, it is pretty easy to show $C_{n+1}$ is countable as well.
For instance, you can say that, since $C_n$ is countable and $\mathbb{Z}$ is countable, also $C_n\times \mathbb{Z}$ is countable, and hence $S:=\{(A,z):A\in C_n,z\in\mathbb{Z}\setminus A\}$ is countable, and we a surjective map from $S$ onto $C_{n+1}$ (sending $(A,z)$ to $A\cup \{z\}$.
Thus, $C_n$ is countable for every $n$.
By the way, a remark: you proved that every basis of this space is countable. This is correct, but you didn't need to show that. You only needed to show that there exists a countable basis of this space.
Best Answer
Take $\Bbb R$ in the discrete metric/topology.
Or consider the Sorgenfrey line, e.g.
Or the lexicographically ordered square. Many exist.