Prove that there is a positive number $B$ such that $|x_{n}| \geq B$ for all $n$ $[{x_n}]^{\infty}_{n=1}$ is a sequence of real numbers that converges to $x_0$ and that all $x_n$ and $x_0$ are nonzero.
I have done the following:
for all $n$ in $|x_{n}|\geq B$
$x_0 < B$
$\epsilon<B-x_o$
$\epsilon+x_o<B$
for all $n$ in $N\geq  n \in  I_{\epsilon}(x_0)=[x_0-\epsilon,x_0+\epsilon]$
$|x_{n}|\geq B$ contradicts $\epsilon+x_o<B$
Would this be a reasonable approach?
I am fairly new to this side of mathematics and have been having trouble with proofs. Introductory Analysis is a pre-requisite to my further studies in Economics which is why I am attempting to learn it.
 A: Hint: There exists $k$ such that $|x_n|>\frac {|x_0|} 2$ for all $n  >k$.  (Choose $k$ such that $|x_n-x_0| <\frac {|x_0|} 2$ for $n >k$). Let  $B$ be the minimum of $\frac {|x_0|} 2$ and $|x_1|,|x_2|,..., |x_k|$. Can you show that $|x_n| \geq B$ for all $n$?
A: Since $x_n$ converges to $x_o$,we have :
For every $\epsilon\gt 0, \exists N_{\epsilon} $ such that for all $n\ge N_{\epsilon} $, we have :
$|x_n-x_o|\lt \epsilon$ 
By triangular inequality, $||x_n|-|x_o||\le |x_n-x_o|$ and hence by choosing $\epsilon =|x_o|/2$,we get:

$||x_n|-|x_o||\lt |x_o|/2 \implies |x_0|-|x_o|/2\lt |x_n|\lt |x_o|+|x_o|/2\implies |x_n|\gt|x_o|/2$ 
Hence, $|x_n|\gt |x_o|/2$ for all $n\ge N_{\epsilon=|x_o|/2}$
Let $B= min\{|x_1|,|x_2|,...,|x_{(N_{\epsilon=|x_o|/2 }-1) }|,|x_o|/2\}$ 
Hence, $|x_n|\ge B$ for all $n\in \mathbb N$
A: We have $x_n \to x_0$ as $n \to \infty$. Then, by definition, for any $\epsilon > 0$,
$$
\exists N \in \mathbb{N} : \forall n \geq N, | x_n - x_0 | < \epsilon
$$
If we choose $\epsilon = |\frac{x_0}{2}|$ we get that for all $n \geq N$
$$
|x_n - x_0| < |\frac{x_0}{2}| \Longrightarrow |x_n| \geq |\frac{x_0}{2}|
$$
If you struggle to see why this is true you can separate it into $x_0 >0$ and $x_0 < 0$ and work it out from there.
Then, you will have that for all $n$,
$$
|x_n| \geq \min \{|x_1|, |x_2|, ..., |x_N|, |\frac{x_0}{2}|\} > 0
$$
A: Since you are learning introductory analysis I will advise you to express the arguments in plain natural language.
Try to grasp this fact

If $x_0$ is non-zero (eg say $1,2,-0.005$ or $10^{-20}$) then it is at a specific distance from $0$ and there is a range of numbers which lie between $0$ and $x_0$. Thus if we get too close to $x_0$ we move far away from $0$.

The sequence in question converges to $x_0$ and thus the terms $x_n$ (after a certain value of $n$, say $m$) can be ensured to lie very close to $x_0$ and thus far away from $0$. Your ask how close to $x_0$? Well, just choose any specific number, say $A$, between $0$ and $x_0$ and then one ensures that the terms $x_n$ for all $n\geq m$ do not lie in the range between $0$ and $A$, but rather between $A$ and $x_0$ or on the other side of $x_0$.
This way we ensure that $|x_n|\geq |A|$ for $n\geq m$. Now choose $B$ to be the minimum of $|A|, |x_1|,|x_2|,\dots,|x_{m-1}|$ (each of these is positive so that $B$ is also positive) and then we have $|x_n|\geq B$ for all $n$.
In such a situation we say that sequence $x_n$ is bounded away from zero.

If you have understood the above you will realize that $\epsilon, \delta$ stuff of analysis is just an unambiguous way to present the same argument. While it does require some experience to convert the informal argument into a formal one, this translation is not the rigor part of analysis and one should not focus too much on it. Rather the focus should be on rigor and key ideas.
