= Solution
<Dynkin lemma>, also called the pi-lambda theorem, states that if a <Dynkin system> $\mathcal D$ contains a <pi-system> $\mathcal P$, then
$$
\sigma(\mathcal P)\subseteq\mathcal D.
$$
Let $\lambda(\mathcal P)$ be the smallest Dynkin system containing $\mathcal P$. It is enough to prove that $\lambda(\mathcal P)$ is a <sigma-algebra>, because then
$$
\sigma(\mathcal P)\subseteq\lambda(\mathcal P)\subseteq\mathcal D.
$$
First fix $A\in\mathcal P$ and define
$$
\mathcal G_A
=\{B\in\lambda(\mathcal P):A\cap B\in\lambda(\mathcal P)\}.
$$
This is a Dynkin system. It contains $\Omega$ because $A\in\lambda(\mathcal P)$; closure under relative complements follows from
$$
A\cap B^c=A\setminus(A\cap B),
$$
and closure under disjoint unions follows by distributing $A\cap-$ over the union. Since $\mathcal P$ is closed under intersections, $\mathcal P\subseteq\mathcal G_A$. Minimality therefore gives
$$
\lambda(\mathcal P)\subseteq\mathcal G_A.
$$
Thus $A\cap B\in\lambda(\mathcal P)$ whenever $A\in\mathcal P$ and $B\in\lambda(\mathcal P)$.
Now fix $B\in\lambda(\mathcal P)$ and define
$$
\mathcal H_B
=\{A\in\lambda(\mathcal P):A\cap B\in\lambda(\mathcal P)\}.
$$
The same argument shows that $\mathcal H_B$ is a Dynkin system, and the preceding paragraph shows that it contains $\mathcal P$. Hence $\mathcal H_B=\lambda(\mathcal P)$. We have proved that $\lambda(\mathcal P)$ is itself a pi-system.
A Dynkin system that is also a pi-system is a sigma-algebra: it is closed under arbitrary finite intersections, hence finite unions by complements, and any countable union can be disjointified before using closure under disjoint unions. Therefore $\lambda(\mathcal P)=\sigma(\mathcal P)$, proving the lemma.
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