Proof verification of $x_n = \sqrt[3]{n^3 + 1} - \sqrt{n^2 - 1}$ is bounded 
Let $n \in \mathbb N$ and:
  $$
x_n = \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1}
$$
  Prove $x_n$ is bounded sequence.

Start with $x_n$:
$$
\begin{align}
x_n &= \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1} = \\
&= n \left(\sqrt[^3]{1 + {1\over n^3}} - \sqrt{1 - {1\over n^2}}\right)
\end{align}
$$
From here:
$$
\sqrt[^3]{1 + {1\over n^3}} \gt 1 \\
\sqrt{1 - {1\over n^2}} \lt 1
$$
Therefore:
$$
\sqrt[^3]{1 + {1\over n^3}} - \sqrt{1 - {1\over n^2}} \gt 0
$$
Which means $x_n \gt 0$.
Consider the following inequality:
$$
\sqrt[^3]{n^3 + 1} \le \sqrt{n^2 + 1} \implies \\
\implies x_n < \sqrt{n^2 + 1} - \sqrt{n^2 - 1}
$$
Or:
$$
x_n < \frac{(\sqrt{n^2 + 1} - \sqrt{n^2 - 1})(\sqrt{n^2 + 1} + \sqrt{n^2 - 1})}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} = \\ 
= \frac{2}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} <2
$$
Also $x_n \gt0$ so finally:
$$
0 < x_n <2
$$
Have i done it the right way?

Your prove is fine but a lot more work than necessary.
As $n \ge 1$ we have
$n = \sqrt[3]{n^3} < \sqrt[3]{n^3 + 1} < \sqrt[3]{n^3 + 3n^2 + 3n + 1} = \sqrt[3]{(n+1)^3} = n+1$
and 
$n = \sqrt{n^2} > \sqrt{n^2 -1 } = \sqrt{n^2 - 2 + 1} \ge \sqrt{n^2 - 2n + 1} = \sqrt{(n-1)^2} = n-1$.
So $0 = n - n < \sqrt[3]{n^3 + 1} - \sqrt{n^2 -1} < (n+1) - (n-1) = 2$.
