Relating to a topology and its basis.

Solution 1:

Note that the collection of finite intersections of the subbasis also include all the subbasis themselves, whereas the collection of all $[a,b[$ does not, so we still need to verify that the two collections generate the same topology. But this is just a routine check.

Let $\mathscr{B}$ be a basis for $\tau$. For each $x\in\mathbb{R}$, pick $B_x\in\mathscr{B}$ such that $x\in B_x\subseteq[x,x+1[$. If $x<y$, then $x\notin B_y$, so $B_x\neq B_y$. This establishes an injection $\mathbb{R}\to\mathscr{B}$, proving $\mathscr{B}$ to be uncountable.

Part (c) is, as you said, not so different from the case of usual topology on $\mathbb{R}$.