Exercise. Show that for the topology $\tau$ in $\mathbb{R}$ with the subbasis sets $A = \{y \in \mathbb{R} : x \leqslant y\}$ and $\mathbb{R} \backslash A = \{y \in \mathbb{R} : x \not\leqslant y\}$as we have the following:
$(a)$ The intervals $[a,b[$ form a basis of $\tau$ closed relatively to binary intersection.
$(b)$ $\tau$ doesn't have a countable base, but for any point it has a neighborhood countable basis.
$(c)$ $\mathbb{Q}$ is dense in $\mathbb{R}$, relatively to $\tau$. (i.e., the closure of $\mathbb{Q}$ is $\mathbb{R}$).
My attempt. I was able to do exercise $(a)$ just by noticing that a basis for the topology $\tau$ is the collection of all finite intersections of elements from the subbasis, i.e., a base $\mathcal{B}$ for the topology $\tau$ is given by: \begin{equation*} \mathcal{B} = \bigcap_{x,x' \in \mathbb{R}} \{A(x),\mathbb{R}\backslash A(x')\} = [a,b[, \quad \text{ for some } a,b \in \mathbb{R}. \end{equation*}
As for exercise $(b)$, how would one proceed? I understand that we must show that every basis for the topology $\tau$ must be uncountable, and I also understand that the definition of a neighborhood countable basis is given by: Let $(X,U)$ be a topological space. We say that a point $x \in X$ has a countable neighborhood base at $x$ if there is a countable collection $\{U^x_j\}^{\infty}_{j=1}$ of open subsets of $X$ such that every neighborhood $W$ of $x$ contains some $U^x_j$, but I don't know how to proceed, unfortunately.
As for exercise $(c)$, what is the difference between proving $\Bbb Q$ is dense in $\Bbb R$ in a general case and proving it relatively to the $\tau$ topology? I feel like nothing changes...
Thanks for all the help in advance.
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 $\mathbb{R}$.