Is this object a set? Or does this break something? Let $S$ be a set-like object that satisfies this three properties:
$1.$ $S$ is not empty.
$2.$ If $a\in S$ then $\{a\}\in S$.
$3.$ If $a_i \in S$ for all $i\in I$ for some indexed set $I$ then $\{a_i\mid i\in I\}\in S$.
This cannot be a set because this looks like every subset of $S$ is also an element of $S$ and so $\mathcal{P}(S) \subseteq S$ (which is not possible due to Cantor's diagonal argument).
What if we change $3$ by
$3'$. If $a_1, a_2 \in S$ and are also sets, then $a_1\cup a_2 \in S$
?
Now only finite subsets of $S$ are guaranteed to be elements of $S$. Is $S$ an actual set?
 A: There are sets satisfying (1), (2), and (3)', so in that sense the answer to your question is yes. However, it's worth noting that those three conditions don't uniquely determine a set, so it's a bit incorrect to say things like "$S$ is an actual set," since there is no specific "$S$."
For example, $V_\alpha$ satisfies your conditions (1), (2), (3') whenever $\alpha$ is a limit ordinal.
A: It is possible to have a set $S$ which satisfies (1), (2), and (3').  Indeed, let $A$ be any nonempty set.  Recursively define $A_0=A$ and $A_{n+1}=A_n\cup\{\{a_1,a_2\}:a_1,a_2\in A_n\}$ for each $n\in\mathbb{N}$.  Then $S=\bigcup_{n\in\mathbb{N}} A_n$ is a set which contains $A$ and satisfies (3'), since for any $a_1,a_2\in S$, there are $m,n\in\mathbb{N}$ such that $a_1\in A_m$ and $a_2\in A_n$, and then $\{a_1,a_2\}\in A_{\max(m,n)+1}\subseteq S$.  It follows that $S$ also satisifes (2) by taking $a_1=a_2=a$.
More generally, a similar construction works for any version of (3) where you put a bound on the cardinality of the indexing set $I$, by iterating this process by a transfinite induction whose length has cofinality greater than $|I|$. For instance, if you wanted $S$ to actually contain all finite subsets of itself, you could do the same construction with $A_{n+1}=A_n\cup\{F:F\subseteq A_n\text{ and $F$ is finite}\}$.  Or if you wanted $S$ to contain all countable subsets of itself, you could do an induction of length $\omega_1$ where $A_{\alpha+1}=A_\alpha\cup\{C:C\subseteq A_\alpha\text{ and $C$ is countable}\}$.
