As the title says, I'm trying to prove that both definitions of a $\lambda$-system are equivalent to each other. It's maybe a bit longer than usual but I hope it's still within the scope of what's OK to ask here.
Dynkin System. A Dynkin System (or $\lambda$-system for short) is a family $\mathcal{F}$ of subsets of $\Omega$ with the following properties:
a) $\Omega\in\mathcal{F}$
b) $A\in\mathcal{F}\implies A^C\in\mathcal{F}$
c) $A_n\in\mathcal{F},n\geq1,A_i\cap A_j=\emptyset, i\neq j\implies\bigcup_{n=1}^\infty A_n\in\mathcal{F}$
An alternative definition for a Dynkin System uses (instead of properties $b$ and $c$):
b') $A,B\in\mathcal{F}, B\subset A\implies A\setminus B\in\mathcal{F}$
c') $A_n\in\mathcal{F},n\geq1,A_n\nearrow\implies\bigcup_{n=1}^\infty A_n\in\mathcal{F}$
Disclaimer: New attempt, old version is deleted as it turned out to be mostly wrong. I am especially concerned about the direction $(\implies)$, more specifically the part $a,b,c\implies c^\prime$.
Proof. $(\Longrightarrow)$ Let us assume that $\mathcal{F}$ is a Dynkin System according to the initial definition using a,b and c. Then iff $B\subset A$ and $A,B\in\mathcal{F}$ we have that $A\setminus B \Leftrightarrow B\cap A^C=\emptyset$ (note that by using b we have ensured that $A^C\in\mathcal{F}$). Applying property c on disjoint sets then lead us to $A\cup B^C\in\mathcal{F}$, thus $A\setminus B\in\mathcal{F}$. Last but not least, let $\Omega\in\mathcal{F}$ and set
$$ B_{n}:=\bigcup_{k=1}^n A_{k}$$
with $A_{k}$ described as in c, namely $A_{i}\cap A_{j}=\emptyset$ for all $i\neq j$. But then $B_k\subset B_{k+1}$, and applying c' on $B_{n}$ then yields the desired result. (Note that $\bigcup_{k\geq1}A_{k}=\bigcup_{k\geq1}B_{k}$.)
Proof. $(\Longleftarrow)$ Let us be given the alternative definition of a $\lambda$-system. Set $A:=\Omega\in\mathcal{F}$ and $B\in\mathcal{F}$. Note that we have assumption that $A\setminus B\in\mathcal{F}$, i.e. $\Omega\setminus B\in\mathcal{F}\Longleftrightarrow B^C\in\mathcal{F}$. Let us be also given $A_k\subset A_{k+1}$ for all $k\geq1$ where $A_k\in\mathcal{F}$ and $A\setminus B\in\mathcal{F}$ as described in c'. Define $B_1:= A_1$ and $B_k:= A_k\setminus A_{k-1}$ for $k\geq2$ recursively. Then,
$$ \overset{c}{\implies}\bigcup_{k=1}^\infty B_k\in\mathcal{F}\overset{Lemma}{=}\bigcup_{k=1}^\infty A_k\in\mathcal{F} $$
which can be proved via induction as shown below (base clause for $n=1$ is sorta obvious):
$$ \bigcup_{k=2}^{n}(A_k\setminus A_{k-1})\cup A_1=\bigcup_{k=1}^n A_k\qquad\text{(induction hypothesis)}, $$
$$ \begin{align*} n\rightarrow n+1: \bigcup_{k=2}^{n+1}(A_k\setminus A_{k-1})\cup A_1&=(A_{n+1}\setminus A_n)\cup \bigcup_{k=2}^n(A_k\setminus A_{k-1})\cup A_1&&\text{(induction hypothesis)}\\ &=(A_{n+1}\setminus A_n)\cup\bigcup_{k=1}^n A_k&&\text{(complement definition)}\\ &=(A_{n+1}\cap A_n^C)\cup\bigcup_{k=1}^n A_k&&\text{(distributive law)}\\ &=\left(A_{n+1}\cup\bigcup_{k=1}^n A_k\right)\cap\left(A_n^C\cup\bigcup_{k=1}^n A_k\right)\\ &=\bigcup_{k=1}^{n+1} A_k\cap\left(A_n^C\cup\bigcup_{k=1}^n A_k\right)&&\underbrace{(A_n^C\cup A_n)\cup\bigcup_{k=1}^{n-1} A_k}_{=\Omega\cup A}=\Omega\\ &=\bigcup_{k=1}^{n+1} A_k \end{align*} $$
Collecting all my comments:
Let $\mathcal{F}$ be a system obeying $a,b,c$.
Fact 0
$\emptyset \in \mathcal{F}$, which follows from $a$ ($\Omega \in \mathcal{F}$) and $b$ ($\emptyset = \Omega \setminus \Omega \in \mathcal{F}$).
Fact 1
If $B_1,\ldots B_N$ is a finite pairwise disjoint family from $\mathcal{F}$, then $\bigcup_{i=1}^N B_i \in \mathcal{F}$. This follows from $c$, which is formulated for infinite sequences, by simply defining $A_n = B_n$ for $1 \le n \le N$ and $B_n = \emptyset$ for $n > N$. Then all $B_n \in \mathcal{F}$ (by Fact 1) and adding $\emptyset$ doesn't change the union so $c$ says that $\bigcup_{i=1}^N B_n = \bigcup_{n \ge 1} A_n \in \mathcal{F}$.
$\mathcal{F}$ obeys $b'$:
Suppose $A,B \in \mathcal{F}$ and $A \subseteq B$. Then $A$ and $B^c \in \mathcal{F}$ (by $b$) are two (pairwise) disjoint sets from $\mathcal{F}$ and so Fact 1 says that $A \cup B^c \in \mathcal{F}$. But then applying $b$ again we see that $$\mathcal{F} \ni (A \cup B^c)^c = A^c \cap (B^c)^c = A^c \cap B = B \setminus A$$ and we have shown $b'$ holds.
$\mathcal{F}$ obeys $c'$:
Let $A_n \in \mathcal{F}, n \ge 1$ be an increasing sequence from $\mathcal{F}$, so $A_n \subseteq A_{n+1}$ for all $n$. Define $B_1 = A_1$ and $B_n = A_n \setminus A_{n-1}$ for $n \ge 2$.
I claim that $\bigcup_{n\ge 1} A_n = \bigcup_{n\ge 1} B_n$ (and this does not use or need induction): let $x \in \bigcup A_n$. Let $m$ be the first index such that $x \in A_m$ (which exists as $\mathbb{N}$ is well ordered, we define $m = \min\{n: n \ge 1, x \in A_n\}$ where the min is taken over a non-empty subset of $\mathbb{N}$). If $m=1$ then $x \in A_1 = B_1$ and otherwise $x \notin A_{m-1}$ by minimality and $x \in A_m \setminus A_{m-1} = B_m$. In both cases $x \in \bigcup_{n \ge 1} B_n$. So $\bigcup_{n \ge 1} A_n \subseteq \bigcup_{n \ge 1} B_n$. The reverse inclusion is trivial as $B_n \subseteq A_n$ for all $n$. QED for the union equality.
As $b'$ holds, all $B_n \in \mathcal{F}$. Also, the $B_n$ are pairwise disjoint: let $m < n$ be two indices. Suppose that $x \in B_m \cap B_n$. Then $x \in A_n$ and $x \notin A_{n-1}$ and also $x \in B_m \subseteq A_m$. But the last implies that also $x \in A_{n-1}$ (as $m \le n-1$ and the $A_n$ are increasing). Contradiction, so $B_n \cap B_m = \emptyset$.
So $c$ implies that $\mathcal{F} \ni \bigcup_{n \ge 1} B_n = \bigcup_{n\ge 1}A_n$ and $c'$ has been shown.
This concludes the $\Rightarrow$ direction of the proof.
Let $\mathcal{F}$ be a system obeying $a,b',c'$.
$\mathcal{F}$ obeys $b$:
Let $A \in \mathcal{F}$. We know $\Omega \in \mathcal{F}$ from $a$. Trivially $A \subseteq \Omega$ so $b'$ says $\mathcal{F} \ni \Omega \setminus A = A^c$. So $b$ holds.
$\mathcal{F}$ obeys $c$:
Let $A_n$ be a pairwise disjoint sequence of sets from $\mathcal{F}$. We define $B_n = \bigcup_{i=1}^n A_i$. It's quite obvious that for all $n$, $B_n \subseteq B_{n+1}$ and that $\bigcup_{n \ge 1} B_n = \bigcup_{n\ge 1} A_n$ so we are done by applying $c'$ if we can show that for all $n$: $B_n \in \mathcal{F}$. For this we can use induction:
$n=1$ is trivial, as $B_1 = A_1 \in \mathcal{F}$.
Suppose that $B_n \in \mathcal{F}$. Then $(A_{n+1})^c \in \mathcal{F}$ by $b$ (which we already showed) and $$B_{n+1}= A_{n+1} \cup B_n = (A_{n+1}^c \cap B_n^c)^c = (A^c_{n+1}\setminus B_n)^c$$ which is in $\mathcal{F}$ by applying $b'$ to $A^c_{n+1}$ and $B_n$ and then applying $b$ again. That $B_n \subseteq A^c_{n+1}$ is clear from the disjointness of the $A_n$ and the definition of $B_n$. This concludes the induction step and so all $B_n \in \mathcal{F}$ and we are done showing $c$.
This concludes the $\Leftarrow$ part of the equivalence of definitions.