Yes, you have the right idea. But the solution you gave is not complete.
Let $\mathfrak{B}$ be the Borel set generated by intervals of the form $[a,b]$ and $\mathfrak{B}^*$ be the Borel set generated by elements of the form $(a,b)$
By the relation you gave,
$$(a,b)=\bigcup_{n=1}^\infty [a+1/n,b-1/n]$$
You showed that any open interval $(a,b)$ can be written as a countable union of closed sets.
Hence, $(a,b)\in \mathfrak{B}$
Now since each element of $\mathfrak{B}^*$ is a countable union, intersection and complement of open intervals. Thus you have
$$\mathfrak{B}^*\subseteq\mathfrak{B}$$
Now, you also need to show $\mathfrak{B}\subseteq\mathfrak{B}^*$.
I'm sure, since you have the right idea, that you'll be able to show that too.
Denote $\mathcal{B}=\sigma\left(\tau\right)=\sigma\left(\left\{ \left(a,b\right)\mid a,b\in\mathbb{R}\right\} \right)$ (this equality was found out by you allready)
where $\tau$ denotes the topology and $\mathcal{B}_{1}=\sigma\left(\left\{ \left(a,\infty\right)\mid a\in\mathbb{R}\right\} \right)$.
Then $\left\{ \left(a,\infty\right)\mid a\in\mathbb{R}\right\} \subset\tau$
so that $\mathcal{B}_{1}\subseteq\mathcal{B}$.
To prove the converse
inclusion it is enough to show that $\left\{ \left(a,b\right)\mid a,b\in\mathbb{R}\right\} \subset\mathcal{B}_{1}$.
We have $\left(a,c\right]=\left(a,\infty\right)-\left(c,\infty\right)\in\mathcal{B}_{1}$
for each $c$, and consequently $\left(a,b\right)=\bigcup_{n\in\mathbb{N}}\left(a,b-\frac{1}{n}\right]\in\mathcal{B}_{1}$.
Best Answer
The mentioned identity shows that the open interval $(a,b)$ can be written as a countably infinite union of half open intervals $(a,\ b-\frac1n]$.
While, $(a,b)$ cannot be an element of the given set, as any finite union of half open intervals has a maximum element, while the open interval doesn't have.
So, the collection of half open intervals is not closed under countable union, i.e. it is not a $\sigma$-algebra.
This in itself doesn't show that, on the other hand, it is an algebra. But that can be easily verified: the union of two elements (finite unions of half open intervals) and the complement of one can be written as a finite union of half open intervals.