If a non-empty, proper subset $A$ of $\mathbb{R}$ is $RR$-open, does this imply $A = (a, \infty)$ for some real number $a$?
Let $RR$ be defined as $$\{ V \subset \mathbb{R}: \text{ if } x \in V,\text{then there exists a ray } (a, \infty) \text{ for some } a \in \mathbb{R} \text{ with } x \in (a, \infty)\subset V \}.$$
My intuition says yes, because if I consider the finite intersection of proper, non-empty elements of $RR$, or the union of a collection of proper, non-empty elements of $RR$, they can never be disjoint. Hence for the union we'd make a ray. Let $$t = \min\big(\text{all the } a_i, \text{with }a_i \text{ being real numbers for each right ray element}\big).$$ Then we have a right ray of $(t, \infty)$.
As well for the finite intersection, let $m = \max\big(\text{of all the }a_i\big)$ to give us a ray $(m, \infty)$. Then yes, $A$ has to be a ray of the form $(a, \infty)$.
But, I do know: consider the usual topology
$$U=\{ D \subset \mathbb{R}: \text{ if } x \in D,\text{ then } x \in (a, b)\subset D\}.$$
If $A$, a non-empty, proper subset of $\mathbb{R}$, is $U$-open, does this imply $V = (a,b)$ for some $a,b \in \mathbb{R}\,$? No, because we could have a $U$-open subset like $(1,2) \cup (2, 3)$ which is $U$-open but not of the form $(a, b)$.
So would the implication in the case of $RR$-openness be true because we can't make any disjoint unions of the elements of the right ray?
Yes, take $p\notin A$ and $q\in A$. Then $p<q$ because $[q,\infty)\subseteq A$. Let $a=\inf A$ (this exists: $A$ is nonempty and bouded below). Then $a\notin A$ because otherwise there would be an $x$ with $(x,\infty)\subseteq A$ and so $A$ would not be the infimum of $A$. But if $x>a$ there there is $y\in A$ with $y<x$ but then $x\in[y,\infty)\subseteq A$. We see $A=(a,\infty)$.