A popular pair of exercises in first courses on functional analysis prove the following theorem:



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> The unit ball of a Banach space $X$ is compact if and only if $X$ is finite-dimensional.
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My question is, is the "only if" part of this (i.e., that the unit ball of an infinite-dimensional Banach space is noncompact) necessarily true without some form of the axiom of choice? The usual proof uses the Hahn--Banach theorem, which may reasonably be regarded as a weak form of the axiom of choice (see [this answer](https://mathoverflow.net/questions/5351/whats-an-example-of-a-space-that-needs-the-hahn-banach-theorem/5566#5566), and other answers to the same question, for some interesting points related to this). 



Is there an infinite-dimensional Banach space with a compact unit ball?
Let me try a possible answer. Take a model of $ZF$ where the axiom of choice for a denumerable family of finite sets holds but where there is an infinite Dedekind finite set $B$ (this model can be checked to exist, for instance, [here](https://web.archive.org/web/20171102221007/http://consequences.emich.edu:80/conseq.htm); $\mathcal{M}32$ is one such model). Then $\ell\_2(B)$ is an infinite dimensional Hilbert space with a Dedekind finite orthonormal base, whose unit ball is, by theorem 2 of Brunner's 1983 article *[Sequential compactness and the axiom of choice](https://doi.org/10.1305/ndjfl/1093870222)*, sequentially compact.


