TITLE: Does ZF imply a weak version of Hahn-Banach?
QUESTION [15 upvotes]: I have encountered this when I was thinking about differentiability in Banach spaces. There, for $x\in X$ we usually need functionals $u\in X^*$ such that $|u|=1$ and $u(x)=|x|$. This is a simple consequence of Hahn-Banach theorem and enables one to convert the problem at hand into a problem in ordinary calculus. Now My question is:
Suppose we have a Banach space $X$ whose dual $X^*$ separates points, i.e. for every nonzero $x\in X$ there is $u\in X^*$ such that $u(x)\neq 0$. Can one prove in ZF that for all nonzero $x\in X$ there is $u\in X^*$ such that $|u|=1$ and $u(x)=|x|$ ?
I know that Hahn-Banach theorem is strictly weaker than axiom of choice, but I'm looking for a proof without using any "choicy" argument.

REPLY [12 votes]: Yemon Choi suggested a counterexample in the comments which can easily be made into an actual counterexample. 
It is consistent with ZF (see Asaf Karagila's answer here for a few references) that the dual space of $\ell_\infty$ is $\ell_1$ with the usual norm. The duality pairing is
$$
\langle x,y \rangle = \sum x_n y_n.
$$
Since the coordinate functionals separate points in $\ell_\infty$, the hypothesis in the question is satisfied.
On the other hand, for the sequence $x_n = 1-1/n$ in $\ell_\infty$ there obviously is no sequence $y \in \ell_1$ of norm one such that  $\sum x_n y_n = 1 = \lVert x_n \rVert_\infty$.