Totally bounded set, $\varepsilon$-nets

I'm working through Martin Schechter's "Principles of Functional Analysis" (2nd ed.) and a problem concerning totally bounded subsets and $\varepsilon$-nets of normed linear vector spaces on page 96 is really bugging me.

I'll begin with the relevant definitions. Throughout, $X$ is a normed linear vector space.

Definition ($\varepsilon$-net) Let $\varepsilon > 0$ be given. A set of points $W \subseteq X$ is called an $\varepsilon$-net for a set $U \subseteq X$ if for every $x \in U$ : $\exists z \in W$ such that $\|x -z \| < \varepsilon$.

Definition (totally bounded) A subset $U \subseteq X$ is called totally bounded if for every $\varepsilon > 0$ there exists a finite set of points $W \subseteq X$ which is an $\varepsilon$-net for $U$.

The following statement in the text is left as an exercise for the reader and it's driving me crazy.

"If $U$ is totally bounded, then for each $\varepsilon >0$ it has a finite $\varepsilon$-net $W \subseteq U$. "

Any ideas?

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Maybe it's too late for me. But isn'T your question trivial? Do you mean it perhaps the other way. Like: $U$ has for every $\varepsilon <0$ a finite $\varepsilon$-net implies $U$ is totally bounded? –  Quickbeam2k1 Mar 4 '13 at 21:12
@Quickbeam2k1: No, it’s not quite trivial: the definition guarantees an $\epsilon$-net in $X$, not in $U$. –  Brian M. Scott Mar 4 '13 at 21:17

Hint: Given $\epsilon > 0$, let $W \subseteq X$ be a finite $\frac{\epsilon}{2}$-net for $U$. For each $z \in W$ pick, if possible, some $x_z \in U$ with $\| x_z - z \| < \frac{\epsilon}{2}$. The triangle inequality should help at this point.