Proof. It suffices to show that the $\s$-algebra generated by any of the above families coincide and the resulting $\s$-algebra is the Borel $\s$-algebra on $\R$. For any $a,b\in\R$
(1)$\to$(2):$(a,\infty)=(-\infty,a)^{c}$
(2)$\to$(3):$[a,\infty)=\caps{}{n\in\N^+}(a-1/n,\infty)$
(3)$\to$(4):$(-\infty,a)=[a,\infty)^{c}$
(4)$\to$(5):$[a,b)=(-\infty,b)\cap(-\infty,a)^{c}$
(5)$\to$(6):$[a,b]=\caps{}{n\in\N^+}(a-1/n,b]$
(6)$\to$(7):$(a,b]=\cups{}{n\in\N^+}[a+1/n,b]$
(7)$\to$(8):$(a,b)=\cups{}{n\in\N^+}(a,b-1/n]$
(8)$\to$(1):$(-\infty,a]=\cups{}{n\in\N^+}\caps{}{k\in\N^+}(-n,a+1/k]$
Here we can always take $n>>1$ sufficiently large. For any $U\suf \R$ open and $q\in U\cap \Q$, there exists $r_{q}>0$ such that $(q-r_{q},q+r_{q})\suf U$. Hence

so (8) generates all open sets in $\R$. Conversely (8) consists of open sets so the $\s$-algebra generated by (8) is the Borel $\s$-algebra on $\R$.$\square$