# Show that $A$ is totally bounded.

Let $(X,d)$ be a metric space and let $A\subseteq X$.

Exercise: Show that if every open cover of $A$ has a finite subcover, then $A$ is totally bounded.

I know that:

For every open cover $C = \bigcup\limits_{n\in \mathbb{N}}C_i$, where $C_i \subset A$ is open and $A \subseteq C$, there exists an open finite subcover $S = \bigcup\limits_{i = 1}^{n}C_i$, such that $A\subseteq S$.

$A$ is totally bounded if for every $\epsilon > 0$ there exists a finite number of points $x_1, \,...,\,x_n \in X$ such that $A\subseteq \bigcup\limits_{i = 1}^{n} B_\epsilon(x_i)$.

Question: How do I show that $A$ is totally bounded? I know that an alternative definition of totally boundedness for A is that every open cover of $A$ has a finite subcover. However, I need to show that $A$ is totally bounded by using the definition I gave above..

Let $\varepsilon > 0$. Notice that $\{B(x, \varepsilon) : x \in A\}$ is an open cover of $A$. Thus, there exists its finite subcover $\{B(x_1, \varepsilon), \ldots, B(x_n, \varepsilon)\}$ for some $x_1, \ldots, x_n \in A$.
Since $\varepsilon$ was arbitrary, $A$ is totally bounded.