I need to prove the following:

Suppose $\displaystyle M$ is a metric space and $\displaystyle A \subseteq M$. Then $\displaystyle A$ is totally bounded if and only if, for every $\displaystyle \epsilon >0$, there is a finite $\displaystyle \epsilon$-dense subset of $\displaystyle A$.

$\displaystyle (\Rightarrow )$: Let $\displaystyle \epsilon > 0$. Then $\displaystyle \epsilon > 0 $. So there exist sets $\displaystyle A_1, A_2, \cdots, A_k$, with $\displaystyle diam(A_i)<\epsilon $ and $\displaystyle A=\displaystyle \cup_{i=1}^{k} A_i$. Now $\displaystyle \forall i$ let $\displaystyle a_i \in A_i$. Let $\displaystyle X=\{ a_i \} _{i=1}^{k}$. So $\displaystyle X$ is a finite subset of $\displaystyle A$. I claim that $\displaystyle X$ is $\displaystyle \epsilon$-dense in $\displaystyle A$. To see this, let $\displaystyle a \in A$. I must show that $\displaystyle \exists$ some element of $\displaystyle X$ so that $\displaystyle \rho (a,x)< \epsilon$. If $\displaystyle a \in A$, then $\displaystyle a \in A_i$ for some $\displaystyle i$. Since $\displaystyle a_i \in A_i$, $\displaystyle diam(A_i) < \epsilon$, $\displaystyle \rho (a, a_i) < \epsilon$.

$\displaystyle (\Leftarrow )$: Now suppose that for $\displaystyle \epsilon > 0$, $\displaystyle A$ has a finite $\displaystyle \epsilon$-dense subset. I must prove that $\displaystyle A$ is totally bounded. Since $\displaystyle A$ has a finite $\displaystyle \epsilon$-dense subset, $\displaystyle \exists \ a \in A$ so that $\displaystyle \rho (x,a)< \epsilon$. So $\displaystyle diam(A)=$ l.u.b.$\displaystyle \{ \rho (a_1, a_2)|a_1, a_2 \in A \}< \epsilon$ so $\displaystyle diam(A)< \epsilon \ \forall a$.

Now for the problem: I am not sure if I have included all the definition for totally bounded; that is, I think I still have another step, but am not sure. Thanks for your help.