I think the following are also sub-$\sigma$-algebras:
$$\sigma(A_1\cup A_2),\quad\sigma(A_1\cup A_3),\quad\sigma(A_1\cup A_4).$$
More generally, let $\{A_1,\ldots,A_n\}$ be a partition of $\Omega$ and let $S=\hat{\sigma}(A_2,A_3,\ldots,A_n)$ be a $\sigma$-algebra of $\Omega\setminus A_1$.
Then $\sigma(A_1\cup B)$ is a sub-$\sigma$-algebra of $\sigma(A_1,\ldots,A_n)$ for all $B\in S$.
To illustrate, let $n=4$ so that
\begin{eqnarray*}
S &=& \hat{\sigma}(A_2,A_3,A_4) \\
&=& \{\emptyset,\;\;A_2,\;\;A_3,\;\;A_4,\;\;A_2\cup A_3,\;\;A_2\cup A_4,\;\;A_3\cup A_4,\;\;A_2\cup A_3\cup A_4 \}.
\end{eqnarray*}
The $\sigma$-sub-algebras of $\sigma(A_1,\ldots,A_4)$ then are:
$$\qquad\sigma(A_1),\quad\sigma(A_1\cup A_2),\quad\sigma(A_1\cup A_3),\quad\sigma(A_1\cup A_4),\quad\sigma(A_1\cup A_2\cup A_3),\quad\sigma(A_1\cup A_2\cup A_4),\quad\sigma(A_1\cup A_3\cup A_4),\quad\sigma(\Omega).$$
I think this would apply also to countable partitions.
A motivation for Dynkin systems (and specially, the $\pi$-$\lambda$ Theorem) is the following problem:
How many sets should I check to be sure that two probability measures $\mu$ and $\nu$ are the same?
For the sake of definiteness, both $\mu$ and $\nu$ are defined on a measurable space $(\Omega,\Sigma)$. A trivial answer to the previous question is “If they coincide over $\Sigma$, they are the same”. True, and useless. So let's think about this for a moment. If I have checked that $\mu(A) =\nu(A)$ for some $A\in\Sigma$, it is not necessary to check for the complement, since
$$
\mu(A^c) =1-\mu(A) = 1-\nu(A) = \nu(A^c).
$$
It is also immediate by $\sigma$-additivity that if they coincide on a sequence of pairwise disjoint sets $A_n$, they must coincide on their union. Hence we conclude
If two probability measures coincide on a family $\mathcal{A}\subseteq\Sigma$, then they coincide on the Dynkin system generated by $\mathcal{A}$ (i.e., the smallest $\lambda$-system containing it).
The above arguments can't be generalized to intersections; we can't compute $\mu(A\cap B)$ from $\mu(A)$ and $\mu(B)$. So, it would be desirable that our initial data (sets checked for coincidence) is a family of sets closed under binary intersections. Here, the $\pi$-$\lambda$ confirms this intuition: If $\Sigma$ is generated by a family $\mathcal{A}$, then it equals the smallest Dynkin system including the closure of $\mathcal{A}$ under intersections (i.e., the $\pi$-system generated by it).
Therefore, to check that $\mu$ and $\nu$ are equal, it is enough to take a family $\mathcal{A}$ such that of $\Sigma =\sigma(\mathcal{A})$ and check that the measures coincide on every finite intersection of members of $\mathcal{A}$.
Best Answer
That condition that partly defines a Dynkin system applies to less sets, because you only have to apply this condition to those that are pairwise disjoint. In particular, if you are a $\sigma$-algebra, then you're a Dynkin system.
A toy example - we say we have a $\sigma$-collection of balls if
we say we have a $\lambda$-collection of balls if
Here, a $\sigma$-collection is always a $\lambda$-collection. But it could be that there are two $30\, \mathrm g$ balls in my $\lambda$-collection, and one costs \$5, and the other costs \$1.
In addition, you can add assumptions to a Dynkin system to turn it into a $\sigma$-algebra. Namely: a Dynkin system $\mathcal D$ that is also closed under finite intersections is a $\sigma$-algebra. That is - suppose $A_i$ is a countable collection of sets. Define $B_1 = A_1$ and inductively $$B_i = A_i \cap A_{i-1}^c \cap \dots \cap A_1^c \in \mathcal D.$$ But for $i<j$, $$B_i \cap B_j \subset A_i \cap B_j = \emptyset$$ so now by the Dynkin property, $$ \bigcup_{i=1}^\infty A_i = \bigcup_{i=1}^\infty B_i \in \mathcal D $$ so $\mathcal D$ is closed under countable unions.
See also : (a) this example that shows that there exist Dynkin systems that are not $\sigma$-algebras, and (b) the $\pi-\lambda$ theorem of Dynkin.