A compact metric space is totally bounded (Alternative proof) To show that a sequentially compact metric space $(X,d)$ is totally bounded is left as an exercise in my textbook, I know that this is very easy if we use: sequentially compact $\Leftrightarrow$ compact, but the book gives a hint to prove it with a different approach. Hint: Assume that $(X,d)$ is not totally bounded and use the following lemma
Lemma: Let $X$ be a set, and let $\Omega\subseteq 2^{X}$ be a collection of subsets of $X$, assume that $\Omega$ does no contain the empty set, then there exist $\Omega'\subseteq\Omega$ such that $A\cap B=\emptyset$ for all $A,B\in\Omega'$ with $A\not=B$ and for all $A\in\Omega$ there exist $B\in\Omega'$ such that $A\cap B\not=\emptyset$.
to find a sequence of balls $B(x_n,\epsilon)$ which are disjoint from each other. I found a similar proof here at the beginning of page 18, but I'm not sure how I have to use the lemma that the exercise gives.
 A: For $\varepsilon > 0$, let $\Omega_\varepsilon$ be the set of $\varepsilon$-radius balls, centred at $x$, as $x$ ranges over $X$. By the lemma, we can find some $\Omega_\varepsilon'$ satisfying the properties in the lemma. This gives us a set of disjoint $\varepsilon$-radius balls. If we can find $\varepsilon$ such that $\Omega'_\varepsilon$ is infinite, then we are done; just choose a countable subset of $\Omega'_\varepsilon$, and we have our sequence of balls.
Since we assume $X$ is not totally bounded, there exists some $\varepsilon > 0$ such that $X$ does not admit a finite $\varepsilon$-net. That is, we assume that, for some $\varepsilon > 0$, there is no finite set $\{x_1, \ldots, x_n\}$ such that
$$X \subseteq \bigcup_{i=1}^n B(x_i; \varepsilon). \tag{$\star$}$$
I claim that, given this $\varepsilon$, $\Omega'_{\varepsilon/2}$ must be infinite.
Why? Suppose contrapositively that $\Omega'_{\varepsilon/2}$ is built from $\Omega_{\varepsilon/2}$ as per the lemma, but
$$\Omega'_{\varepsilon/2} = \{B(x_1; \varepsilon/2), \ldots, B(x_n; \varepsilon/2)\}.$$
Suppose $x \in X$. Then $B(x; \varepsilon/2) \in \Omega_{\varepsilon/2}$. Thus, from our construction of $\Omega'_{\varepsilon/2}$, there exists some $i = 1, \ldots, n$ such that
$$B(x; \varepsilon/2) \cap B(x_i; \varepsilon/2) \neq \emptyset.$$
Let $y$ be an element of the above set. Then,
$$d(x, x_i) \le d(x, y) + d(y, x_i) \le \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$
(or strict inequality, depending on whether you mean open or closed balls). Either way, $(\star)$ holds true, as required.
