For concreteness' sake, let the underlying set theory be ZFC.
A set $x$ will be called connex (I've adapted the terminology from here) iff for every $y, z \in x$ either $y \subseteq z$ or $z \subseteq y$.
The following theorem characterizes finite sets:
A set $x$ is finite iff for every non-empty connex set $y$ whose union is $x$, $x \in y$.
To see this, let $x$ be a set.
Suppose that $x$ is finite, and let $y$ be a non-empty connex set such that $x = \cup y$. If $x$ is empty, it must be that $y = \{\emptyset\}$, and indeed $x \in y$. If $x$ is non-empty, for every $a \in x$ choose a set $f(a) \in y$ such that $a \in f(a)$. It suffices to show that $x \in f[x]$. Choose any $a_1 \in x$. If $f(a_1) = x$, we're done. Otherwise, choose any $a_2 \in x\setminus f(a_1)$. By connexity, it must be that $f(a_1) \subseteq f(a_2)$, and therefore that $\{a_1,a_2\} \subseteq f(a_2)$. If $f(a_2) = x$, we're done. Otherwise, proceed in this fashion. After at most $|x|$ steps, we will obtain $x$.
Conversely, suppose that $x$ is infinite. Let $s$ be an infinitely countable subset of $x$, and let $s_1, s_2, \dots$ be a bijection from $\{1,2,\dots\}$ on $s$. Define $y_0 := x\setminus s$, and, for every $n \in \{1,2,\dots\}$, $y_n := y_{n-1}\cup\{s_n\}$. Observe that, for every $n \in \{0,1,2,\dots\}$, $y_n \subsetneq y_{n+1} \subseteq x$. Then $y = \{y_0, y_1, y_2, \dots\}$ is a non-empty connex set whose union is $x$, however $x \notin y$, since otherwise we would have $x = y_n$ for some $n \in \{0,1,2,\dots\}$, but then $x \subsetneq y_{n+1} \subseteq x$!
QED
Along similar lines, we can make the following definition for every cardinality $c$.
A set $x$ will be called $c$-finite iff for every non-empty connex set $y$ of cardinality $c$ whose union is $x$, $x \in y$.
It can be shown that a set is $\aleph_0$-finite iff it is finite. One direction (namely $\Leftarrow$) is immediate from the characterization of finiteness given above. As for the other direction, let $x$ be an infinite set, and define $y = \{y_0, y_1, y_2, \dots\}$ as in the above proof. Then $y$ is a connex set of cardinality $\aleph_0$ whose union is $x$, however $x \notin y$, as was shown above.
My questions
Is the concept of $c$-finiteness known, perhaps under a different name or with some variation? Is there a simpler way of saying that a set $x$ is $c$-finite? In particular, is the following true?
A set is $c$-finite iff its cardinality is strictly less than $c$.
The Dedekind cut argument showing that $\aleph_0$ is not $2^{\aleph_0}$-finite is universal, in a precise sense: the question of $\kappa$-finiteness amounts to counting Dedekind cuts (for somewhat vacuous reasons).
First, let's put a bound on the possible cardinalities of connexes, right off the bat. Trivially there are no connexes with union $x$ of cardinality $>2^{\vert x\vert}$, but it turns out we can do better. For a cardinal $\mu$, let $\overline{Ded}(\mu)$ (not standard notation, sadly) denote the smallest cardinal which is larger than the number of Dedekind cuts in any linear order of size $\mu$. For example, we have $\overline{Ded}(\aleph_0)=2^{\aleph_0}$ by considering the rationals. In general, it turns out that we can have $\overline{Ded}(\mu)<2^\mu$ (see e.g. here).
Now suppose $Y$ is a connex whose union is $x$. We get an induced preorder $\triangleleft$ on $x$, given by setting $a\triangleleft b$ iff for each $y\in Y$ we have $b\in y\implies a\in y$. Elements of $Y$ correspond injectively (since connexes are sets, rather than sequences, of sets) to Dedekind cuts in this preorder (or in a linear order refining this preorder, if you prefer). So there are no connexes with union $x$ of cardinality $\ge\overline{Ded}(\vert x\vert)$, and hence $x$ is vacuously $\kappa$-finite for every $\kappa\ge\overline{Ded}(\vert x\vert)$.
In the other direction, suppose $x$ is infinite and there is a linear order $L$ of size $\vert x\vert$ with no maximal element, cofinality $\lambda$, and $\kappa$-many Dedekind cuts. Every "unbounded" set of Dedekind cuts in $L$ yields a connex whose union is $x$ but which doesn't contain $x$ as an element, and we can find such a set of any cardinality between $\lambda$ and $\kappa$ inclusive. Moreover, we can always ensure $\lambda=\omega$ (just consider $L+\omega$ ...), so this just says that every set $x$ is $\mu$-infinite whenever $\mu$ is infinite and $<\overline{Ded}(\vert x\vert)$.
Finally, every set is trivially (and annoyingly) $n$-finite for every finite $n$. Putting this all together, we have: