TITLE: Lemma 2.5-2 and Theorem 2.5-3 in Kreyszig's Functional Analysis Book: Does compactness of every closed and bounded subset also imply ...?
QUESTION [5 upvotes]: Here is Lemma 2.5-2 in the book Introductory Functional Analysis With Applications by Erwine Kreyszig:

A compact subset $M$ of a metric space is closed and bounded. 

But the converse is not true, as is shown by the set $$M = \left\{ \ (1, 0, 0, \ldots), \ (0, 1, 0, 0, \ldots), \ (0, 0, 1, 0, 0, \ldots), \ \ldots \ \right\}$$ in $\ell^2$. 
But here is Theorem 2.5-3: 

In a finite dimensional normed space $X$, any subset $M \subset X$ is compact if and only if $M$ is closed and bounded.

Now my question is as follows: 

Let $X$ be a metric space such that every closed and bounded subset of $X$ is (sequentially) compact. Does this imply that $X$ is a finite dimensional normed space?

I have no idea of how to proceed, although I'm clear about the proofs in Kreyszig.

REPLY [2 votes]: This is not true, and the basic reason is that most metric spaces are not normed vector spaces at all.  For instance, no normed vector space of positive dimension can be compact.  So if $X$ is any compact metric space at all with $|X|\neq 1$, $X$ cannot be homeomorphic to any normed vector space, but any closed subset of $X$ is compact.
In fact, it is possible that $X$ is not even homeomorphic to a subset of any finite dimensional normed vector space.  For instance, let $X$ be the Hilbert cube, which is a compact metric space.  Then $\mathbb{R}^n$ embeds topologically in $X$ for all $n$, since $X$ is homeomorphic to $[0,1]^\mathbb{N}$.  It follows that $X$ cannot embed topologically in $\mathbb{R}^N$ for any $N$, since $\mathbb{R}^n$ only embeds in $\mathbb{R}^N$ if $n\leq N$ (this is a hard theorem in topology; it follows from invariance of domain, for instance).
(Incidentally, you should really be more precise about what you mean by "$X$ is a finite dimensional normed space".  Do you mean it is isometric to a finite dimensional normed space, or homeomorphic, or something else?  However, I think the examples above show pretty conclusively the answer will be no for any reasonable interpretation.)