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My incorrect thought was to find uncountably many uniformly separated points. Now suppose countably many totally bounded sets $\{B_n\}$ cover these points, it is then apparent that each $B_n$ covers at most finitely many of them since each sequence in a totally bounded set has a Cauchy subsequence, and hence the contradiction.

Unfortunately there are a great many separable Banach spaces (eg. $\ell^2$), which invalidates my "proof". So I request an alternative way. And, as a side note, if it wouldn't complicate matters very much, please avoid using the fact that an infinite-dimensional Banach space admits an uncountable Hamel basis.

Best regards!

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A metric space is compact if and only if it is complete and totally bounded, so a $\sigma$-totally bounded Banach space is $\sigma$-compact and hence finite-dimensional.

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  • $\begingroup$ Thanks, this already solves my problem. Yet as a followup question, is there any easy way without appeal to BCT (in proving $\sigma$-compact $\Leftrightarrow$ finite-dimensional in particular)? $\endgroup$ – Vim Oct 10 '16 at 16:43
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    $\begingroup$ @Vim: You’re welcome. I don’t know, but I rather suspect that there is not. $\endgroup$ – Brian M. Scott Oct 10 '16 at 16:47

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