Every set is a union of singleton sets Is it possible to prove this statement? What axioms are necessary to conclude that every set on, for instance, $\mathbb{R}$ can be expressed as a union of singleton sets (sets containing exactly one element)?
Let $X=\{x : P\}$ where $P$ is a defining property of the elements of $X$, for instance $P$ can be $x>2$.
Then I can simply state that $$X = \bigcup_{P} X_i$$ where $\{X_i : i \in \mathbb{R_+}\}$ is a collection of all singleton sets of $\mathbb{R}$.
By the way, there are sets of bigger cardinality than of cardinality of all real numbers. In this case, using $\mathbb{R_+}$ as indexes for all singleton sets from such sets wouldn't be sufficient, right?

First of all, given two sets $A$ and $B$ we say that
$$\tag{1}
A=B
\quad\Longleftrightarrow\quad
A\subseteq B
~\text{ and }~
B\subseteq A
$$
That is, the sets coincide if every element of $A$ is in $B$ and every element of $B$ is in $A$.
As @MSE says, the intuition is that
$$\tag{2}
  M = \bigcup_{x\in M}\{x\}
\qquad\text{(to be proven)}
$$
To prove this, we shall show that both inclusions hold.
It is convenient to define
$$\tag{3}
\mathscr M := \bigl\{\{x\}:~x\in M\bigr\}
$$
and rewrite $(2)$ as
$$\tag{4}
  M = \bigcup_{S\in \mathscr M}S
\qquad\text{(equivalent to $(2)$, to be proven)}
$$
Now, let $x\in M$, then $\{x\}\in\mathscr M$ by definition $(3)$, and therefore
$
\{x\}\subseteq\bigcup_{S\in\mathscr M}S
$,
which implies
$
x\in\bigcup_{S\in\mathscr M}S
$.
Since $x\in M$ is arbitrary, this proves that
$$
  M \subseteq \bigcup_{S\in \mathscr M}S
$$
Conversely, let $x\in\bigcup_{S\in \mathscr M}S$, then $x\in S$ for some $S\in\mathscr M$.
By definition $(3)$ of $\mathscr M$, there exists $y\in M$ such that $S=\{y\}$.
Since $x\in\{y\}$, it follows that $x=y\in M$.
Again, from the arbitrarity of $x\in\bigcup_{S\in \mathscr M}S$ it follows that the other inclusion holds as well
$$
  \bigcup_{S\in \mathscr M}S \subseteq M
$$
This proves the sought identity $(4)$ and hence $(2)$.
