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.

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 $n$,
$$
|x_n| \geq \min \{|x_1|, |x_2|, ..., |x_N|, |\frac{x_0}{2}|\} > 0
$$
