Borel algebra is generated by the collection of all half-open intervals

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.

You showed that the $\sigma$-algebra generated by the closed (resp. half-open) intervals is a superset of the Borel algebra. The other inclusion is still missing: Obtain $[a,b]$ using nothing but countable unions and complements of open intervals. A straight-forward possiblity is $$[a,b]=\mathbb R\setminus\left(\bigcup_{n=1}^\infty\left(\mathbb R\setminus(a-1/n,b+1/n)\right)\right)\text.$$

Reusing this representation, we obtain $$(a,b]=(a,(a+b)/2)\cup[(a+b)/2,b]$$ for half-open intervals.