Instead of just re-posting my first comment as an answer, I thought it would be better to comment briefly on your original question. (As I said in my second comment, your solution in the edit works fine).
Unfortunately (or rather fortunately) your original approach doesn't work as you intended. Here are three simple observations:
- Every closed set in a metrizable space is a $G_{\delta}$. In particular every point is a $G_{\delta}$.
- Recall the following form of the Baire category theorem: "the intersection of countably many dense $G_{\delta}$s in a complete metric space is again a dense $G_{\delta}$". This immediately implies that a countable dense subset of a perfect Polish space (e.g. $\mathbb{Q}$ in $\mathbb{R}$) is an $F_{\sigma}$ that isn't a $G_{\delta}$.
- It is obvious from the second sentence in 1. that a countable dense set is a countable disjoint union of $G_{\delta}$'s. But as I argued, it won't be a $G_{\delta}$ itself in general.
Finally, let me just note that the detour I suggested in my first comment is not strictly necessary, as by unpacking the proof of the equivalent descriptions you should be able to see what to do in order to save your original direct approach. However, it seems rather obfuscating than clarifying what is going on, so I leave that to you in case you're interested.
Note that if $A_n$ is any family of sets, then
$$ \bigcup_{n\in\mathbb N}A_n = \bigcup_{n\in\mathbb N}\bigl(A_n\setminus\bigcup_{k=0}^{n-1} A_k\bigr)$$
where the summands on the right-hand side are disjoint, and each of them is constructed from finitely many of the $A_i$s by a sequence of complements and finite unions.
So if you choose you can restrict yourself to requiring countable unions of disjoint set, if you also require that the algebra is closed under finite unions of arbitrary sets.
A Dynkin system that is not a $\sigma$-algebra:
If you don't require arbitrary finite unions, what you get is not necessarily a $\sigma$-algebra. Consider for example the system of subsets of $\mathbb R$ consisting of $\varnothing$, $\{0,x\}$ for every $x\ne 0$, and the complements of these sets. It is closed under your proposed axioms, because the only nontrivial disjoint union there is to take is $\{0,x\} \cup \{0,x\}^\complement = \mathbb R$.
A small finite example is
$$ \bigl\{\varnothing, \{0,2\}, \{0,3\}, \{1,2\}, \{1, 3\}, \{0,1,2,3\} \bigr\} $$
Best Answer
An algebra of sets $\mathcal{A}$ on a set $X$ is closed under finite unions and complements, so also under finite intersections and relative complements (via De Morgan's laws). The algebra $\mathcal{A}$ is said to be a $\sigma$-algebra if it is closed under countable unions.
In particular, a $\sigma$-algebra is closed under “countable disjoint unions” (that is unions of countable families consisting of pairwise disjoint members of $\mathcal{A}$).
Is the converse true? Let's assume the algebra $\mathcal{A}$ is closed under countable disjoint unions and consider any countable family $(A_n)_{n\ge0}$.
We can define a new family by $$ B_{n}=A_n\setminus\bigcup_{k<n}A_k= A_{n}\setminus(A_0\cup A_1\cup\dots\cup A_{n-1}) $$ In particular $B_0=A_0$ (because we're subtracting the empty set) and $B_n\subseteq A_n$. Suppose $m<n$; then we can show that $B_m\cap B_n=\emptyset$. Indeed if $x\in B_m$ we have $x\in A_m$, but as $m<n$ we see that $x\in A_0\cup\dots\cup A_{n-1}$ and so $x\notin B_n$.
Moreover, we clearly have $$ \bigcup_{n\ge0}B_n\subseteq \bigcup_{n\ge0}A_n $$ Suppose $x\in\bigcup_{n\ge0}A_n$. Thus, for some $n$, we have $x\in A_n$ and we can take $n$ to be minimal with respect to this property. Thus $x\notin A_m$ for $m<n$ and therefore $x\in B_n$. So we have proved that $$ \bigcup_{n\ge0}A_n=\bigcup_{n\ge0}B_n\in\mathcal{A} $$ because we assumed closure under countable disjoint unions.