Show that if $(x_n)$ is unbounded, then there exist a subsequence $\left(x_{n_{k}}\right)$ such that $\lim(\frac{1}{x_{n_{k}}}) = 0$.
By the monotone converge theorem if $X$ converges there exists a subsequence that's either increasing or decreasing.
Case 1 (increasing): Since $X$ is increasing $x_{n_{1}}$ > $x_{n_{2}}$ > ... > $x_{n_{k}}$. Since $X$ is unbounded its subsequence is also unbounded, and therefore approaches infinity. Therefore the reciprocal $\frac{1}{x_{n_{1}}}$ > $\frac{1}{x_{n_{2}}}$ > ... > $\frac{1}{x_{n_{k}}}$ goes to $0$.
Case 2 (decreasing): Since $X$ is decreasing $x_{n_{1}}$ < $x_{n_{2}}$ < ... < $x_{n_{k}}$. Since $X$ is unbounded its subsequence is also unbounded, and therefore approaches negative infinity. Therefore the reciprocal $\frac{1}{x_{n_{1}}}$ > $\frac{1}{x_{n_{2}}}$ > ... > $\frac{1}{x_{n_{k}}}$ goes to $0$.
Is this proof sufficient? Is there a better way to show it, or do it?
I'd argue like this.
Since $(x_n)$ is unbounded, there has to be some term $x_{n_1}$ so that $| x_{n_1} | \geq 2$. Inductively suppose we've picked the indices $n_1, n_2, \cdots, n_{k-1}$. Given $k$, again because $(x_n)$ is unbounded, we can find an $n_k > n_{k-1}$ so that
$$ \left\vert x_{n_k} \right\vert \geq 2^k $$
Now if you want to show formally that
$$ \lim_{k \to \infty} \dfrac{1}{x_{n_k}} = 0 $$
then for any $\varepsilon > 0$, pick $K$ such that $2^{-K} < \varepsilon$, and note that
$$ \begin{align*} | 1 / x_{n_k} | & \leq 2^{-k} \\ & \leq 2^{-K} \\ & < \varepsilon \end{align*} $$
for all $k \geq K$.