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
$$ \bigl\{\varnothing, \{0,2\}, \{0,3\}, \{1,2\}, \{1, 3\}, \{0,1,2,3\} \bigr\} $$
Let $$\mathscr C\equiv\{E\subseteq\mathbb R\,|\,E\text{ is countable or }E^c\text{ is countable}\}.$$
It is not difficult to check that $\mathscr C$ is a $\sigma$-algebra on $\mathbb R$. Let $\mathscr S$ denote the $\sigma$-algebra generated by singletons.
$\textbf{Proposition:}\phantom{---}$$\mathscr C=\mathscr S$.
$\textit{Proof:}\phantom{---}$ Clearly, $\{x\}\in\mathscr C$ for every $x\in\mathbb R$, so $\mathscr S\subseteq\mathscr C$, since $\mathscr C$ is a $\sigma$-algebra containing the singletons and $\mathscr S$ is, by definition, the smallest $\sigma$-algebra containing the singletons. Conversely, suppose that $E\in\mathscr C$. If $E$ is countable, then $E$ is a countable union of singletons. Therefore, $E\in\mathscr S$ (because $\mathscr S$ is $\sigma$-algebra containing the singletons). If $E^c$ is countable, then, by the same argument, $E^c\in\mathscr S$, from which it follows that $E\in\mathscr S$ (since $\sigma$-algebras are closed under complements). In conclusion, $\mathscr C\subseteq\mathscr S$. $\blacksquare$
Clearly, $(0,1)\notin\mathscr C=\mathscr S$, but $(0,1)\in\mathscr B_{\mathbb R}$. It follows that $\mathscr B_{\mathbb R}\neq\mathscr S$.
Best Answer
Let $\mathcal{S}=\{\{x\}:x\in X\}$ be the family of all singletons of $X$ and $\sigma(\mathcal{S})$ the $\sigma$-algebra generated by $\mathcal{S}$. Since $S$ is $\sigma$-algebra and $\mathcal{S}\subset S$, we have $\sigma(\mathcal{S})\subset S$. Consequently, it only remains to prove that $S\subset\sigma(\mathcal{S})$. Let $A\in S$. If $A$ is countable, then $A\in \sigma(\mathcal{S})$, due to $A=\bigcup_{x\in A}\{x\}$. If $A$ is not countable, then $A^c$ is countable and hence $A^c\in \sigma(\mathcal{S})$. But $\sigma(\mathcal{S})$ is $\sigma$-algebra and, accordingly, $A\in \sigma(\mathcal{S})$. Therefore, in any case $A\in \sigma(\mathcal{S})$, which proves that $S=\sigma(\mathcal{S})$.