Solution Verification: Convergence of Norms $\implies$ Convergence of Sequence I found the following exercise in Bartle's Elements of Real Analysis. I'm learning on my own and have a couple of doubts would love it if someone could take a look. ($\left|{\left|{x}\right|}\right|$ represents the standard Euclidean norm).  

Let $X = (x_n)$ be a sequence in $\Bbb R^p$ and let $\lim ( \left|{
 \left|{x_n }\right|}\right| ) = 0$. Show that $\lim (x_n) = 0$.
  However give an example in $\Bbb R$ to show that the convergence of
  $(\left|{x_n}\right|)$ may not imply the convergence of $(x_n)$

This is my attempt:
Solution Part 1: $$\left|{ \left|{ x_n - 0 }\right| }\right| = \left|{\left|{x_n }\right|}\right| = \left|{\left|{x_n }\right|}\right| - 0 = \left| \ {\left|{\left| { x_n }\right|}\right| - 0} \ \right| $$ Therefore given any $\epsilon \gt 0$ there is an  $m \in \Bbb N$ such that $n \ge m \implies \left|{ \left|{ x_n - 0 }\right| }\right| = \left| \ {\left|{\left| { x_n }\right|}\right| - 0} \ \right| \lt \epsilon $. Therefore $\lim (x_n) = 0$. 
Solution Part 2: The counter-example I came up with was $(a_n) = \left({ (-1)^n \left[{ 3 + \frac 1 n}\right]  }\right)$. Since $\left|{a_n}\right| = \left|{3 + \frac 1 n}\right| = 3 + \frac 1 n $, it can be proven that $\lim (\left|{a_n}\right|) = 3$. Now to prove that $(a_n)$ diverges: 
Suppose it converges to a limit $l \in \Bbb R$. Let $\epsilon = 1 \gt 0 $. Then there is $m \in \Bbb N$ such that $m \ge n \implies \left|{a_n - l}\right| \lt 1$. Let $m_1$ and $m_2$ be two respectively odd and even natural numbers greater than or equal to $m$. Then, 
$$ \left|{a_{m_1} - l}\right|, \left|{a_{m_2} - l}\right| \lt 1 \implies \left|{\left ( -\frac 1 {m_1} - 3\right) - l}\right| \lt 1 \;\; \text { and } \;\; \left|{\left(3 + \frac 1 {m_2}\right) - l}\right| \lt 1 $$ 
Adding the two inequalities we have $$ \left|{ 2l - \left({ \frac 1 {m_2} - \frac 1 {m_1}}\right) }\right| \lt 2 \implies 2 \left|{l }\right| \lt 2 + \frac 1 {m_1} + \frac {1}{m_2} \lt 2 + \frac 2 m \implies \left|{l}\right| \lt 1 + \frac  1 m$$
Now say we pick $\epsilon = \frac 1 {m}$. Then, 
$$\left|{a_n - l}\right| \ge \left|{a_n}\right| - \left|{l}\right| \gt \left|{a_n}\right| - \left(1 + \frac 1 m\right) \gt 3 - \left(1 + \frac 1 m\right)  = 2 - \frac 1 m \ge \frac 1 m  \;\;\; \forall n \in \Bbb N \rightarrow\leftarrow $$
I have a couple of doubts and would greatly appreciate it if someone could help me out:
$Q_1:$ Well I'm not sure how to prove the divergence of a sequence so am not sure about the proof of the non-existence of the limit in Part 2. Is it good enough? Is there a better way? If so guide me..
$Q_2:$ Part 1 does not stipulate the fact that $p \gt 1$. It generally shows for any Cartesian space $\Bbb R^p$ the convergence of the norms of the elements in the sequence implies convergence of the sequence. Then how come we have a counter-example in $\Bbb R$??
 A: Q1. It is an alternative counterexample: take $a_n=(-1)^n$, then we have $\|x_n\|\to 1$ but $\{x_n\}$ is divergent. Why? Assume $\{x_n\}$ is convergent to $x$. Put $\epsilon=\frac{\|1-x\|+\|1+x\|}{2}>0$. Since $x_n$ is convergent then there exists $N$ such that $\|x_n-x\|<\epsilon$ for all $n\geq N$. However, taking $n=2N$ and $n=2N+1$ gives $\|x_{2N}-x\|=\|1-x\|$ while  $\|x_{2N+1}-x\|=\|-1-x\|=\|1+x\|$ that is impossible both of them less that $\epsilon$
. So, $\{x_n\}$ is divergent.
Q2. In the Part1, the sequence has to be converged to $0$. If not, the implication will not work.
A: This counter-example for part 2 in the above question is courtesy @Jlamprong but I think I have an easier way to prove its divergence and I'm posting it as an answer mainly for the community to check if its valid. 
Let $(a_n)$ be a sequence in $\Bbb R$ defined by $a_n= (-1)^n$. Suppose it converges to a limit $a$. Let $\epsilon = \frac 1 2. $ Then there is $m \in \Bbb N$ such that $n \ge m \implies \left|{ a_n - a }\right| \lt \frac 1 2.$ Without loss of generality assume $m$ is even. Then $\left|{a_m - a}\right| = \left|{1 - a}\right| \lt \frac 1 2 \implies a \in ( \frac 1 2 , \frac 3 2 )$. But then; $$\left|{a_{m + 1} - a}\right| = \left|{-1 -a}\right| = \left|{a + 1}\right| = a + 1 \gt \frac 1 2 \Rightarrow\Leftarrow$$
