= Solution
A set is <totally bounded> if for every $\varepsilon>0$ it can be covered by finitely many open balls of radius $\varepsilon$.
Here is the needed separated-sequence argument, including the relevant <Riesz lemma>. For a proper <closed subspace> $Y$, choose $z\notin Y$, put $d=\operatorname{dist}(z,Y)>0$, and choose $y\in Y$ with $\|z-y\|<2d$. Then $w=(z-y)/\|z-y\|$ is a <unit vector> at distance greater than $1/2$ from $Y$. Starting with one <unit vector>, apply this to the span of the vectors already chosen. These spans are closed and proper because they are finite-dimensional while $X$ is infinite-dimensional. The resulting unit vectors satisfy $\|w_m-w_n\|>1/2$ for $m\ne n$. A ball of radius $1/4$ contains at most one of them. Hence \b[the <closed unit ball> is not <totally bounded>].
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