let $\displaystyle x_n $ be an unbounded sequence . prove that there is a subsequence $\displaystyle x_{n_k} $ of $\displaystyle x_n $ such that $\displaystyle \frac{1}{x_{n_k}} \to 0 $
Suppose that $\displaystyle \left(x_n\right)$ is not bounded above.
Then $\displaystyle \left( {\exists x_{N_1 } } \right)\left[ {1 < x_{N_1 } } \right]$.
Let $\displaystyle B_1 = \max \left\{ {2,N_1 } \right\}$ then $\displaystyle \left( {\exists x_{N_2 } } \right)\left[ {B_1 < x_{N_2 } } \right]$.
Now you can see how to construct the required subsequence.