How does one prove $ZF\vDash MC\Rightarrow AC$? This is somewhat adressed to Andreas Blass, whose papers I have read, in particular I make reference to an old paper of his »Existence of Basis implies the Axiom of Choice« (84). Anyone who happens to be informed may nevertheless feel free to enlighten me.

One works within $ZF$. Assume $MC$, that is, for all $X$ there exists $f:\mathcal{P}(X)\to\mathcal{P}(X)$, so that for all $A\in\mathcal{P}(X)\setminus\{\emptyset\}$ it holds $\emptyset\neq f(A)\subseteq A$ and $f(A)$ finite. To show is $AC$.
Due to $MC$ it holds
Proposition 0. Let $X$ be a set. There is a choice function $\mathcal{P}(X)\to X$ $\Longleftarrow$ $X$ is linear-orderable.
Proof. Consider a l.o. $\preceq$ on $X$. Set $m:\mathcal{P}(X)\to\mathcal{P}(X)$ a multi-choice function. Then $c:\mathcal{P}(X)\setminus\{\emptyset\}\to X$ defined by $c(A)=\min_{\preceq}m(A)$ is well-defined and a choice function. $\blacksquare$
Since one is working within $ZF$ it also holds:
Proposition 1. Let $X$ be a set. There is a choice function $\mathcal{P}(X)\to X$ $\Longleftrightarrow$ $X$ is well-orderable. $\dashv$
Corollary. $X$ has choice $\Longrightarrow$ $X$ has a w.o. $\Longrightarrow$ $X$ has a l.o. $\Longrightarrow$ $X$ has choice. $\dashv$
It is thus necessary and sufficient to show, every set is linear-orderable. Again, since one is working within $ZF$, it is necessary and sufficient to just consider the sets $V_{\alpha}$ for $\alpha\in NO$ (the ordinal numbers).
An argument per induction seems to suggest itself. For the successor stages, one simply builds out of a well-order $\preceq$ on $V_{\alpha-1}$ a linear order (the lexical order, $\preceq_{lex}$) on $V_{\alpha}=\mathcal{P}(V_{\alpha-1})\equiv_{SET}{}^{V_{\alpha-1}}2$.
The argument encounters difficulty for limit-ordinals, since one cannot „choose“ a well-order for all the stages below it (except when the ordinal satisfies $cf(\alpha)<\alpha$). Consider simply $\alpha=\omega_{1}$, how would one construct a linear-order on $V_{\omega_{1}}$?

I should note: all authors I have read up until now each hold that $ZF\vDash MC\Rightarrow AC$.

Final Remark. I see now how to handle with successor stages in Frunobulax’ argument.
Thm. Assuming $ZF+MC$ every $V_{\alpha}$ is well-orderable.
Proof. Consider $\alpha\in NO$ any ordinal. Thanks to $MC$ there exists $m:\mathcal{P}(V_{\alpha})\to\mathcal{P}(V_{\alpha})$ a multi-choice function. This shall henceforth remain fixed.
As already mention it holds
Claim. for $X\subseteq V_{\alpha}$ and an LO $\preceq$ on $X$, using just $m$ and $\preceq$ one can canonically construct a choice function $\mathcal{P}(X)\to X$ (via. $c:A\mapsto \min_{\preceq}(m(A))$ for $A\neq\emptyset$) and out of this in turn canonically construct a WO of $X$ (utilising the recursion $x_{\gamma}:=c(X\setminus\{x_{\xi}:\xi<\gamma\})$ for $\gamma\in NO$ for as long as $X\supsetneq\{x_{\xi}:\xi<\gamma\}$). This canonically-definable well-order shall be denoted $\preceq^{m}$. $\dashv$
Construction of a WO of $V_{\alpha}$. One constructs recursively an increasing chain $(\preceq_{\beta})_{\beta\leq\alpha}$ of WO-s of the $V_{\beta}$:


*

*$\preceq_{0}:=\emptyset$.

*Limit stages: $\preceq_{\beta}:=\bigcup_{\xi<\beta}\preceq_{\xi}$.

*Successor stages: one linearly orders $R_{\beta}:=V_{\beta}\setminus V_{\beta-1}\subseteq \mathcal{P}(V_{\beta-1})\equiv{}^{V_{\beta-1}}2$ via the lexical-ordering $(\preceq_{beta-1})_{lex}$. Note that $R_{\beta}\subseteq V_{\alpha}$. Set $\preceq^{\ast}:=(\preceq_{beta-1})_{lex}^{m}$ a well-order. The well-orders are then patched together by placing everything in $R_{\beta}$ after $V_{\beta-1}$, to obtain a well-order of $V_{\beta-1}\sqcup R_{\beta}=V_{\beta}$.


This recursion is well-define and in particular $\preceq_{\alpha}$ witnesses the well-orderability of $V_{\alpha}$. $\blacksquare$
 A: If that's OK with you, it's easy to show that $\textbf{MC}$ implies
the antichain principle which is known to be equivalent to the axiom
of choice:
Let $X$ be a set partially ordered by $R$, i.e. $R$ is reflexive, antisymmetric and transitive.  Let $f$ be a function
which assigns to each non-empty subset $A$ of $X$ a finite non-empty
subset $f(A)$ of $A$.  For each such $A$ the set
of $R$-minimal elements of $f(A)$ is finite,
non-empty and an $R$-antichain.  We'll call it $A'$.
Now, inductively, define $C_\alpha$ to be the set of all elements of
$X$ which are $R$-incomparable with all elements of $B_\beta$ for all
$\beta<\alpha$ and let $B_\alpha$ be $C_\alpha'$ (or $\emptyset$ if
$C_\alpha$ is empty).
The union of all $B_\alpha$ is a maximal $R$-antichain.
A: Here's a modified version of a proof by Rubin which builds on what you
already have above:
We claim that $\textbf{MC}$ implies that all $\mathbf{V}_\alpha$ can
be well-ordered.
So, let $\alpha$ be some fixed arbitrary ordinal.  Let $\lambda$ be
the smallest ordinal which can't be mapped one-to-one into
$\mathbf{V}_\alpha$ (i.e. the Hartogs number of $\mathbf{V}_\alpha$).
$\lambda$ is well-ordered by $\in$, so we can define a linear order
$\preccurlyeq$ on ${\cal P}(\lambda)$ by $x
\preccurlyeq y$ iff $x=y$ or $x\neq y$ and $\min(x\bigtriangleup
y)\in x$ (where $\bigtriangleup$ is the symmetric difference).
As you've shown above, due to $\textbf{MC}$ we'll find a choice
function for ${\cal P}(\lambda)$ which in turn implies that this set
can be well-ordered.  We fix one well-ordering
of ${\cal P}(\lambda)$ for the rest of the proof.
Now let $R_\beta$ be the set of all $x\in{\mathbf V}_\alpha$ with rank
$\beta$.
If we can show, that all $R_\beta$ for $\beta<\alpha$ can be
well-ordered, then it is obvious how we can combine these
well-orderings to get one for ${\mathbf V}_\alpha$.  So, we'll
inductively define well-orderings $\preccurlyeq_\beta$ of $R_\beta$:
Pick $\beta<\alpha$ and assume that for $\gamma<\beta$ a well-ordering
$\preccurlyeq_\gamma$ of $R_\gamma$ has already been defined.  Using
these, we can define an "obvious" well-ordering $<_\beta$ of ${\mathbf
  V}_\beta$ (being the disjoint union of the $R_\gamma$ for
$\gamma<\beta$).
There's a unique ordinal $\delta_\beta$ such that
$(\mathbf{V}_\beta,<_\beta)$ is isomorphic to $(\delta_\beta,\in)$ and
correspondingly there's a unique isomorphism $g_\beta$ from
$(\delta_\beta,\in)$ onto $(\mathbf{V}_\beta,<_\beta)$.  As $g_\beta$
is also an injection from $\delta_\beta$ into $\mathbf{V}_\alpha$, we
must have $\delta_\beta<\lambda$.
Now, for an element $x$ of $R_\beta$ we have $x\subseteq{\mathbf
  V}_\beta$ and thus $g_\beta^{-1}[x] \subseteq \delta_\beta \subseteq
\lambda$.  Thus, $g_\beta$ (or rather its inverse) gives us a natural
way of "projecting" a part of the above-mentioned well-ordering of ${\cal P}(\lambda)$
to $R_\beta$ which finishes our proof.
The important thing to note here is that we didn't make a choice at
each $\beta<\alpha$ stage.  The only choice made was right at the
beginning when we fixed a well-ordering of ${\cal P}(\lambda)$.
