Over ZF, does "every Hilbert space have a basis" imply AC? I know there is a similar result due to Blass [1] that over ZF, "every vector space has a (Hamel) basis" implies AC. Looking around, however, I can't find any results on the question for Hilbert spaces. I also don't see how to generalize Blass's proof to my question.
To be clear, what I am asking is:
Question Over ZF, does "every Hilbert space have a basis" imply AC, where a Hilbert space is a complete inner product space and a basis for a Hilbert space is a set of orthonormal elements whose span is dense?

[1] A. Blass, "Existence of bases implies the axiom of choice", Contempory Mathematics, Vol. 31, (1984).
 A: According to the book Albrecht Pietsch: History of Banach Spaces and Linear Operators (Birkhäuser, 2007, 
DOI: 10.1007/978-0-8176-4596-0) this is an open problem. 
(Or at least it was at the time of publishing the book.)
I quote from p.586:

We list two more consequences of the axiom of choice, which have already been discussed in 1.2.2 and 1.5.10, respectively:
  $(\mathsf{B}_{\mathsf{alg}})$ Every real or complex linear space has a Hamel basis.
  $(\mathsf{B}_{\mathsf{orth}})$ Every real or complex Hilbert space has an orthonormal basis.  
It is unknown whether $(\mathsf{B}_{\mathsf{alg}}) \overset{?}\Rightarrow (\mathsf{AC})$ or $(\mathsf{B}_{\mathsf{orth}}) \overset{?}\Rightarrow (\mathsf{AC})$.
  Some partial results were obtained by Bleicher [1964] and Halpern [1966].
  Blass [1984, p. 31] showed that the axiom of choice follows if we assume that every linear space over an arbitrary field has a basis.



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*J.D. Halpern [1966] Bases in vector spaces and the axiom of choice, Proc. Amer. Math. Soc. 17, 670–673. DOI: 10.1090/S0002-9939-1966-0194340-1

*M.N. Bleicher [1964] Some theorems on vector spaces and the axiom of choice, Fund. Math. 54, 95–107. eudml, matwbn

*A. Blass [1984] Existence of basis implies the axiom of choice, Contemp. Math. 31, 31–33. author's website
