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..
Thanks in advance!