Is the lower limit topology finer than the standard topology on $\mathbb{R}$?

Yes! Since one have that $$ (a,b) = \cup_{n\ge 1} \ [a+\frac{\epsilon}{n},b) $$ where $\epsilon < \frac{b-a}{2}$.

Note that if for topology ${\mathcal T}_1$ with basis ${\mathcal S}_1$ and topology ${\mathcal T}_2$, one have that ${\mathcal S}_1 \subseteq {\mathcal T}_2$, then ${\mathcal T}_2$ is finer than ${\mathcal T}_1$. In this case if ${\mathcal S}_2$ be a basis for topology ${\mathcal T}_2$ and ${\mathcal S}_2 \not\subseteq{\mathcal T}_1$, then ${\mathcal T}_2$ is strictly finer than ${\mathcal T}_1$.For this, it is enough to note that $[0,1)$ is not open in standard topology.

For more details you can consult this textbook: James Munkres, "Topology; A First Course".

Well, I don't have a copy of Munkres' book at hand but I doubt that it is said.

If $[a,b)$ is open then $(a,b)=\bigcup_{n\in\mathbb{N}}\left[ a+\frac{1}{n},b\right)$ must be open.

Conversely, $[0,1)$ is not open in the standard topology.

This means that the topology in $\mathbb{R}_l$ is finer that the topology on $\mathbb{R}.$