Let $\mathbb{R}_{std}$ be the standard topology on $\mathbb{R}$. Let $\mathbb{R}_{LL}$ be the lower limit topology, whose basis elements consists of all intervals of the form $[a,b)$. Prove that every open set in $\mathbb{R}_{std}$ is open in $\mathbb{R}_{LL}$, but not every open set in $\mathbb{R}_{LL}$ is open in $\mathbb{R}_{std}$.
Attempt: For any open set $U$ in $\mathbb{R}_{std}$ for each $x \in U$, there is an open interval $(a,b)$ containing $x$. Then $[x,b)$ is in $\mathbb{R}_{LL}$ and $[x,b) \subset (a,b) \subset U$, so $U$ is open in $\mathbb{R}_{LL}$.
In $\mathbb{R}_{LL}$, however for any $[x,b)$ there is no open interval $(c,d)$ containing $x$ with $(c,d) \subset [x,b)$ thus not every open set in $\mathbb{R}_{LL}$ is open in $\mathbb{R}_{std}$.
Best Answer
Yes, this is the right way. The way you saw is also correct: if basic Euclidean sets (like open intervals) are all LL-open, then so are all Euclidean open sets. This is perfectly rigorous.
But the final argument why $[x,b)$ is not Euclidean open can be slightly expanded: suppose $x \in (c,d) \subseteq [x,b)$ then we can pick $r$ with $c < r < x$ (we know that $c < x < d$ after all, and $\Bbb R$ is a dense order) but then the inclusion tells us that $r \in [x,b)$ so $r \ge x$ while $r < x$, contradiction.
It's clear from the "picture", but a proof shouldn't rely on that only.