Question about Hopf-Rinow theorem I'm studying Hopf-Rinow theorem and I don't see a step in the proof. Could someone help me, please?
(Definition) Let's $(M, \langle,\rangle)$ an ANII(axiom numerability 2) and Hausdorff Riemannian manifold. If $M$ is connected and $p,q \in M$. We define: $$d_L:M\times M \longrightarrow [0,\infty)$$ $$(p,q)\longmapsto inf\{l(C)\}$$
where C is a piecewise differentiable curve joining $p$ and $q$ and $l$ is the length of the curve.
I've proved that $d_L$ is a distance and that the topology induced by $d_L$ is the original topology on $M$.
Considering  $(M, \langle,\rangle)$ an ANII and Hausdorff Riemannian manifold. If $M$ is connected and $p \in M$. Then prove that the following statements are equivalent:
(a) If $A$ is a closed and bounded subset of $M$ then $A$ is compact.
(b) $\exists \{K_n\}_{n\in \mathbb{N}}, K_n \subset M$ compact and $K_n \subset K_{n+1}, \forall n \in \mathbb{N}$ and $\bigcup_n K_n=M$ with the following property:
If $(q_n)_{ n \in \mathbb{N}}\subset M$ sequence / $(q_n)\notin K_n, \forall n \in \mathbb{N}\Rightarrow lim_{n\rightarrow \infty} d_L(q_n,p)=\infty$.
Thanks.
 A: $(\Rightarrow)$ Assume (a) is true. Fix a point $o\in M$ and let $K_n$ be the closed unit ball of radius $n$. That is
$$K_n = \{ x\in M: d(x, o)\leq n\}\ .$$
Then each $K_n$ is closed and bounded, thus they are all compact by (a). Let $y\in M$. Then $d(o, y) \leq n$ for some $n$, so $y\in K_n$. Thus $\bigcup_n K_n = M$. 
Let $p\in M$. Then as $q_n \notin K_n$, $d(o, q_n)> n$. By triangle inequality, 
$$n< d(o , q_n) \leq d(p, o) + d(p, q_n) \Rightarrow d(p, q_n) > n-d(o, p) \to \infty\ ,$$
hence $\lim_n d(p, q_n) = \infty$. 
$(\Leftarrow)$ Assume that $(b)$ is true and $A$ is closed and bounded. As $A$ is bounded, there is $p\in M$ and $C>0$ such that $d(x, p \leq C$ for all $x\in A$. First we show that there is $N\in \mathbb N$ such that 
$$A \subset K_N\ .$$
Assume the contrary that such a $N$ does not exist. Then for all $n\in\mathbb N$, there is $q_n\in A$ such that $q_n \notin K_n$. By (b), we have $d(p, q_n) \to \infty$. But that contradicts the fact that $A$ is bounded. 
Thus $A\subset K_N$ for some $N$. As $K_N$ is compact and $A$ is closed, $A$ is also compact. 
Thus we have shown that (a) and (b) are equivalent. Note that it has nothing to do with manifold. That is true for any metric space.
