Show that the topology generated by $\{$ $(a,\infty)$,$(-\infty,b)$ $:$ $a,b\in \mathbb{R} \}$, $\tau'$, is the standard metric topology, $\tau$ on $\mathbb{R}$
My attempt:
If $U$ is in $\tau'$ then clearly, it contains an open interval, either $(a,\infty)$ or $(-\infty, b)$, both of which are open in the standard metric topology. Hence $\tau' \subseteq \tau$. If $U$ is in $\tau$. Then, for each $x\in U$, there exists $r>0$, for which, $(x-r,x+r)\subseteq U$. But, $(x-r,x+r)=(-\infty,x+r)\cap (x-r,\infty) \subseteq U$. Since these two intervals are in the basis, they must be in the topology it generates. Hence $U\in \tau'$.
Is my attempt correct?
I think it is better to argue on the level of sets, rather than on the level of elements. This gives clearer/cleaner arguments.
Notice that the given subbasis is contained in $\mathcal{T}$ and it immediately follows that $\mathcal{T}' \subseteq \mathcal{T}$
Taking intersections of subbasis elements, we see that $\mathcal{T}'$ contains all intervals of the form $(a,b)$ with $a< b$. Since every $\mathcal{T}$-open set is the union of such intervals, it follows that $\mathcal{T} \subseteq \mathcal{T}'$.