Borel sigma algebra $\mathcal{B} (\mathbb{\bar{R}}) $ generated by $ (a, \infty], a\in \mathbb{R}.$ So I know that the Borel sigma-algebra of  $\mathbb{\bar{R}} $ is generated by open sets in $\mathbb{\bar{R}}$.
I will write:
 $\mathcal{B} (\mathbb{\bar{R}}) $ = $\sigma$ ( { open sets in $\mathbb{\bar{R}} $}).
Now I have to show that:
 $\mathcal{B} (\mathbb{\bar{R}}) $ = $\sigma$ ( { $(a, \infty]$ |  $ a\in \mathbb{R}$}) .
''$\supset$'' is okay.
''$\subset$'' 
I would start with the fact: $(a,\infty]^c$ = $[-\infty,a] \in$ $\sigma$ ( { $(a, \infty]$ |  $ a\in \mathbb{R}$}) .
Then {$a$} $\in$ $\sigma$ ( { $(a, \infty]$ |  $ a\in \mathbb{R}$}). ( because {$a$} = ( $\bigcap_{j=1} ( a -\frac{1}{j},\infty]) \cap [-\infty,a]) $
Then $[-\infty,a)$ $\in$ $\sigma$ ( { $(a, \infty]$ |  $ a\in \mathbb{R}$}) (because $[-\infty,a) = [-\infty,a]$ \ {$a$}) 
Then $(a,b)$ $\in$ $\sigma$ ( { $(a, \infty]$ |  $ a\in \mathbb{R}$}) (because $[-\infty,b) \cap (a,\infty] = (a,b) )$
This means
$\sigma$ ( { $(a, \infty]$ |  $ a\in \mathbb{R}$}) $\supset$ {open sets in $\mathbb{\bar{R}}$}
Now we conclude: $\sigma$ ( { $(a, \infty]$ |  $ a\in \mathbb{R}$}) $\supset$ $\sigma$({open sets in $\mathbb{\bar{R}}$}) = $\mathcal{B} (\mathbb{\bar{R}}) $.
Is this okay? Please look very closely.
 A: You're confusing 'open set' and (bounded) 'open interval'. An open interval is a set of the form $\{x\in\mathbb R\colon a<x<b\}$ for some $a,b\in \mathbb R$, which we symbolize as $(a,b)$.
Remember that a set $A$ is open in $\bar{\mathbb{R}}$ iff for every $a\in A$ there exists an open interval $I$ such that $a\in I \subset A$; or said in another way: $A$ is open in $\bar{\mathbb{R}}$ iff for every $a\in A$ there exists some $\delta>0$ such that $x\in A$ for every $x\in (a-\delta,a+\delta)$.
What you can prove and then use in your proof is that any open set can be expressed as a countable union of bounded open intervals.
Is a little technical but not very complicated: part of the argument is that $\mathbb Q$ is dense in $\mathbb R$ and that $\mathbb Q$ is countable.
EDIT
Let me try once more:
You proved:
$(a,b) \in \sigma \Big( \big\{ (a, \infty] \subset \bar{\mathbb{R}} |   a\in \mathbb{R} \big\}\Big).$
This proves that all open bounded intervals (that is, all "$(a,b)$") are in the $\sigma$-algebra generated by the intervals $(a,\infty]$, and then it is also true that the whole $\sigma$-algebra generated by those intervals $(a,b)$ is also included in $\sigma \Big( \big\{ (a, \infty] \subset \bar{\mathbb{R}} |   a\in \mathbb{R} \big\}\Big)$.
But you said instead that the $\sigma$-algebra generated by the open sets is included in $\sigma \Big( \big\{ (a, \infty] \subset \bar{\mathbb{R}}|   a\in \mathbb{R} \big\}\Big)$.
You could have got to that conclusion if you had proven that all the open sets (not just the intervals $(a,b)$) are included in $\sigma \Big( \big\{ (a, \infty] \subset \bar{\mathbb{R}} |   a\in \mathbb{R} \big\}\Big)$.
It is actually true that all these $\sigma$-algebras we've mentioned are the same, but that is far from obvious. You proved that the intervals $(a,b)$ and the intervals $(a,\infty]$ generate the same $\sigma$-algebra, but since they are all open sets of $\bar{\mathbb{R}}$ it is immediate that the $\sigma$-algebra we just mentioned is a subset of the one generated by the open sets (the Borel $\sigma$-algebra), but the converse is more complicated, since it involves proving that all the open sets (no matter how complicated) are part of the $\sigma$-algebra generated by open bounded intervals or by left open infinite intervals.
And when I say "no matter how complicated", I'm thinking that
$$(0,1)\cup (2,3) \cup (4,5) \cup \ldots,$$
$$\mathbb{R}\setminus \mathbb{Z},$$
$$\left\{x\in \mathbb{R}\colon \sin\left(\frac1{1+x^2}\right)<e^{-x^4}\right\}$$
or
$$\{x\in (0,\infty) \colon \exists k \in \mathbb{Z}(2k-1<\log_2(x) <2k)\}$$
are examples of open sets.
