$B=\{[a,b): a<b\}$ is the basis inducing the topology $T_B$. We want to prove that If $K\subset\mathbb{R}$ is compact w.r.t $T_B$, then $K$ is countable.
My attempt
Given $K$ is compact, then for $x\in K$. we can find the open cover $[x,+\infty) \cup_{n\in \mathbb{N}}(-\infty,x-\frac{1}{n} )$. By compactness, this has a finite subcover, namely,
$[x,+\infty) \cup_{i=1}^{n}(-\infty,x-\frac{1}{i} )=[x,+\infty)\cup(-\infty.x-\frac{1}{n})$
Then for any $x_i\in K$, $\exists n_i \in \mathbb{N}$, such that the corresponding cover includes $K$.
We can also construct another open cover, so that $\cup_{i=1}^{n}[x_i,x_i+\delta)$ is the finite subcover, so we can say its boundedness as well?
This seems relatable to countability, but I do not know what to do next.