Principles of lesser choice in ZF Working in $\mathbf{ZF}$ consider the follows statements:

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*Statement A: Let $X$ be a set that has members of arbitrarily large finite size.  Then $X$ has a countable subset $Y\subseteq X$ with the same property.  (note - members not subsets.)

*Statement B: Let $F$ be a countable set of non-empty mutually disjoint finite sets.  There is a choice set  $C$ such that for each $x\in F$, $C\cap x$ chooses a single member of $x$.

*Statement C: Let $D_0\subseteq D_1\subseteq D_2\subseteq\dots$ be an increasing sequence of finite sets.  Then $\bigcup_{n=0}^\infty D_n$ is countable.

Now, $A\Rightarrow B$, $B\iff C$ and Countable AC implies $A$.  Here is a sketch of proofs:
$C\Rightarrow B$:
Enumerate $F$ and then construct an increasing union of its members as in statement $C$.  The union is countable so can be well-ordered.  Then use a well-order on $\bigcup F$ to do the selection for the choice set.
$B\Rightarrow C$:
What we want to do here is construct by recursion functions $f_n:D_n\rightarrow |D_n|$ such that $f_{n+1}\restriction D_n=f_n$.  Then take $\bigcup_{n=0}^\infty f_n$.  To do this we need to simultaneously choose bijections $b_n:D_{n+1}\setminus D_n\rightarrow |D_{n+1}\setminus D_n|$ for all $n$, and then use the addition function on natural numbers to push the image of $b_n$ to its place in $\omega$.  The number of bijections from a set of size $k$ to a set of size $k$ is $k!$, and each $D_{n+1}\setminus D_n$ is finite, so we can choose all the bijections $b_n$ simulatneously using the tool in statement $B$.
$A\Rightarrow B$:
I assume $|F|=\aleph_0$ since for finite $F$ statement B can be proved by induction on the size of $F$ in $\mathbf{ZF}$ alone.  Well-order $F$ in order type $\omega$, and let $G$ be the set of partial choice sets on $F$ that perform selection on a proper initial-segment of $F$.  The members of $G$ are all finite.  By induction on size of members of $G$, $G$ can be shown to have members of arbitrarily large finite size (this is the same as the case of finite $F$.)  By statement $A$, $G$ has a countable subset $H$ with the same property.  Now well-order $H$.  Then each $x\in F$ is covered by some choice set in $H$, because the initial segment $\le x$ in the well-order of $F$ is finite, and $H$ has partial choice sets on arbitrarily large proper initial segments of $F$.  So for $x\in F$ take the least choice set in the order of $H$ that covers it - and use that one to choose a member of $x$.  This process constructs a choice set for all of $F$.
Countable AC implies statement A:
For $n\in\omega$ let $X_n$ be the set of all members of $X$ of finite size $\ge n$.  Use Countable AC to get a choice function $f:\omega\to X$ such that $f(n)\in X_n$ for all $n$.  Then the image of $f$ is a countable subset of $X$ with the same property.
Now, statements $A$ and $B$ together with $\mathbf{ZF}$ can prove that any infinite set is Dedekind infinite (has a countable subset).  See my answer here for how to do this.  Since statement $A$ implies statement $B$, it follows that $\mathbf{ZF}+A$ implies that any infinite set is Dedekind infinite.  Since there are models of $\mathbf{ZF}$ with infinite sets that are Dedekind finite it follows that statement $A$ is independent of $\mathbf{ZF}$.  My questions are:

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*Is it known whether statement $B$ (or $C$) is independent from $\mathbf{ZF}$?

*Is it known whether statement $A$ is independent from $B$?

*Is it known whether Countable AC is independent from $A$?

 A: The first two have a positive answer, the third one is negative (both that it isn't known, and also that it isn't true).
Let's start with the third one. Statement A is equivalent to $\sf AC_\omega$.
Now suppose that Statement A holds, let $\{A_n\mid n<\omega\}$ be a family of non-empty sets. Let $F_n=\prod_{i<n}A_i$ and let $F=\bigcup F_n$, then $F$ has arbitrarily large finite members, by Statement A it has a countable subset $Y$ with the same property. Note that if $f\in F$ and $|f|>n$, then $f(n)\in A_n$ is defined. Now order $Y$ as $f_n$ such that the sizes are not decreasing, and by whittling down we may assume that the sizes are strictly increasing and $|f_0|>0$. Therefore $|f_n|>n$ for all $n$, and we can define $f(n)=f_n(n)$ as a choice function.
Your statements B and C are imply, for example, that every countable sequence of pairwise disjoint pairs admits a choice function. This is well-known to be unprovable in $\sf ZF$, indeed Cohen's second model is an example of that. This principle is something referred to as $\sf AC_\omega^{\rm fin}$ (do note some authors flip the sub- and superscript). And it is also equivalent to the strong version of Kőnig's lemma: every finite branching tree has an infinite path. (The weak version talks about countable trees, it is provable in $\sf ZF$ but comes up in reverse mathematics.)
Finally, note that B and C follow from "every set can be linearly ordered", since we can fix a linear ordering of the union of these finite sets, and this will uniformly well-order all of the finite sets. In Cohen's first model every set can be linearly ordered, but there are infinite Dedekind-finite sets, so countable choice, and therefore Statement A both fail.
