How to prove this $\mathcal{G}$ is a $\pi$-system? Let $\mathcal{K}$ be a $\pi$-system and let $\mathcal{G}$ be the smallest Dynkin system such that $\mathcal{K} \subseteq \mathcal{G}$. Then $\mathcal{G}=\sigma(\mathcal{K})$.
The proof is: since $\sigma(\mathcal{K})$ is a sigma algebra, thus it is also a Dynkin system. We then obtain $\mathcal{G} \subseteq  \sigma(\mathcal{K})$
In order to prove the opposite inclusion, I notice that if I can say $\mathcal{G}$ is a $\pi$-system, then $\mathcal{G}$ will also be a sigma algebra, thus I am done. But how to prove $\mathcal{G}$ is a $\pi$-system?
One proof is Let $A \in \mathcal{G}$ and define
$$\mathcal{G}_A=\{ B \in \mathcal{G}: A \cap B \in \mathcal{G} \}$$.
The collection of sets $\mathcal{G}_A$ is a Dynkin system, which I understand and know how to prove. But what next?
 A: Let $\delta(.)$ be the smallest Dynkin system associated to a family of sets. We are claiming that if $\mathscr{G}$ is $\cap$-stable, then $\delta(\mathscr{G})=\sigma(\mathscr{G})$. Since $\delta(\sigma(\mathscr{G}))=\sigma(\mathscr{G})$, we obtain $\delta(\mathscr{G})\subseteq \sigma(\mathscr{G})$. Consider $B \in \mathscr{G}$;
$$\mathscr{G}\subseteq \mathscr{H}_B:=\{A\in \delta(\mathscr{G}):A\cap B\in \delta(\mathscr{G})\}\subseteq \delta(\mathscr{G})$$
We may prove $\mathscr{H}_B$ is Dynkin. Since $B \in \mathscr{G}$, then $X \in \mathscr{H}_B$. Let $C \in \mathscr{H}_B$. Then $C\cap B\in \delta(\mathscr{G})$, so $C^c\cup B^c\in \delta(\mathscr{G})$. But then
$$C^c\cap B=(C^c\cup B^c)\cap B=((C\cap B)\cup B^c)^c \in \delta(\mathscr{G})$$
Now let $(C_n)_{n \in \mathbb{N}}\subseteq \mathscr{H}_B$ mutually disjoint. Then
$$(\cup_nC_n)\cap B=\cup_n(C_n \cap B)\in \delta(\mathscr{G})$$
So we conclude $\mathscr{H}_B=\delta(\mathscr{G}),\,\forall B \in \mathscr{G}$. Now, again
$$\mathscr{G}\subseteq \mathscr{C}:=\{B \in \delta(\mathscr{G}):\mathscr{H}_B=\delta(\mathscr{G})\}\subseteq \delta(\mathscr{G})$$
We prove $\mathscr{C}$ is Dynkin. Clearly $X \in \mathscr{C}$. Now let $B \in \mathscr{C}$. Then $A\cap B\in \delta(\mathscr{G}),\,\forall A \in \delta(\mathscr{G})$. Now
$$A\cap B^c=(A^c\cup B^c)\cap A=((A\cap B) \cup A^c)^c\in \delta(\mathscr{G}),\,\forall A \in \delta(\mathscr{G})$$
So $B^c \in \mathscr{C}$. Now let $(B_n)_{n \in \mathbb{N}}\subseteq \mathscr{C}$, mutually disjoint. Then $B_n\cap A \in \delta(\mathscr{G}),\,\forall A \in \delta(\mathscr{G}),\forall n$. We get
$$(\cup_nB_n)\cap A=\cup_n(B_n \cap A)\in \delta(\mathscr{G}),\,\forall A \in \delta(\mathscr{G})$$
So we conclude $\mathscr{C}=\delta(\mathscr{G})$. But this means that $\delta(\mathscr{G})$ is $\cap$-stable. We then prove that $\delta(\mathscr{G})$ is a $\sigma$-algebra. Let $(A_n)_{n \in \mathbb{N}}\subseteq \delta(\mathscr{G})$. Then
$$\bigcup_{n \in \mathbb{N}}A_n=\bigcup_{n \in \mathbb{N}}\underbrace{\bigg(A_n/\bigcup_{k<n}A_k\bigg)}_{\textrm{mutually disj.}}=\bigcup_{n \in \mathbb{N}}\bigg(A_n\cap \bigcap_{k<n}A_k^c\bigg)\in \delta(\mathscr{G})$$
So we conclude $\sigma(\mathscr{G})\subseteq \sigma(\delta(\mathscr{G}))=\delta(\mathscr{G})$.
