 Solution

A set is totally bounded if for every $\epsilon >0$ it can be covered by finitely many open balls of radius $\epsilon$.

Here is the needed separated-sequence argument, including the relevant Riesz lemma. For a proper closed subspace $Y$, choose $z\notin Y$, put $d=\mathrm{dist}(z,Y)>0$, and choose $y\in Y$ with $\Vert z-y\Vert <2d$. Then $w=(z-y)/\Vert z-y\Vert$ 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 $\Vert w_m-w_n\Vert >1/2$ for $m\ne n$. A ball of radius $1/4$ contains at most one of them. Hence the closed unit ball is not totally bounded.