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.



Does ZF imply a weak version of Hahn-Banach?
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](https://math.stackexchange.com/q/103476) 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 $\sum x\_n y\_n = 1 = \lVert x\_n \rVert\_\infty$.


