Show that the topology is strictly finer 
My attempt:
Since the standard topology is the topology whose open sets are the unions of sets of the type $(a,b)\cap X$, with $a,b\in\mathbb R$ and $a<b$ (if $X\subset\mathbb R$).
And since $(a,b) \subset [a,b)$ we can see that $\mathscr T =\cup_{i\in I}(a_i,b) \subset \cup_{i\in I}[a_i,b)=\mathscr T_1$
Would this be correct?
 A: You seem to have misunderstood what needs to be done. To show that $\mathscr{T}_\ell$ is strictly finer than $\mathscr{T}$, you must show that $\mathscr{T}_\ell\supsetneqq\mathscr{T}$. That is, you must show that every standard open set is still open in the topology $\mathscr{T}_\ell$ and that there is at least one open set in $\mathscr{T}_\ell$ that is not a standard open set.
To show that $\mathscr{T}_\ell$ is finer than $\mathscr{T}$, but not necessarily strictly finer, you need only show that $\mathscr{T}\subseteq\mathscr{T}_\ell$. Every standard open set is a union of open intervals, so if you can show that $(a,b)\in\mathscr{T}_\ell$ whenever $a,b\in\Bbb R$ with $a<b$, then you’ll have shown that $\mathscr{T}\subseteq\mathscr{T}_\ell$. You can do this by showing that $(a,b)$ is union of intervals of the form $[x,y)$.
To show that $\mathscr{T}_\ell$ is strictly finer than $\mathscr{T}$, you need to find a set that is in $\mathscr{T}_\ell$ but is not open in the standard topology. The base for $\mathscr{T}_\ell$ gives you some very obvious possibilities.
