Let $X$ by non empty and $D$ a $\delta$-ring. I must prove the set:
$$\sigma(D)=\{A:\exists(A_n)_{n\in \mathbb N}\subset D \hspace{0.2cm} s.t\hspace{0.2cm} A=\bigcup\limits_{i=0}^{\infty} A_{i}\}\tag{1} $$ is the $\sigma$-ring generated by $D$. To do that, I need to prove: (1): $D\subset \sigma(D) $; (2): $\sigma(D)$ is a ring ; (3): $\forall (S_n)_{n\in \mathbb N} \subset\sigma(D)\hspace{0.4cm}\bigcup\limits_{i=0}^{\infty} S_{i}\in\sigma(D)$.
I can prove (1) and (2) but (3) is the problem. I've tried to construct sequence: $S_n=\bigcup\limits_{i=0}^{\infty} A_{i}^{n}$ and then $\bigcup S_n=\bigcup\limits_{n=0}^{\infty}\bigcup\limits_{i=0}^{\infty}A_{i}^{n}$ and if I can write this like a single union of set's from $D$, I will be done. (Edit) And to complete I need to prove $\sigma(D)$ is the smallest in the sense of inclusion but I don't know how.
For 3 you are done, because it's easy to write a countable union of countable unions as a single countable union by reindexing (using some bijection between $\mathbb{N}$ and $\mathbb{N}\times \mathbb{N}$, of which there are plenty).
Minimality is clear: if $\mathcal{S}$ is any $\sigma$-ring that contains $D$, $\sigma(D)$ as defined by your first formula is a subset of $\mathcal{S}$. So it must be the minimal $\sigma$-ring generated by $D$ which justifies the notation.