If the sequence is bounded, then you can invoke Bolzano-Weierstrass theorem.
So we can assume it is unbounded. Let us assume it is not bounded from above. (The symmetric case is similar.)
Thus for every $C$ there exists $n$ such that $x_n>C$.
Start with $C_1=1$ and choose $n_1$ such that $x_{n_1}>C_1=1$.
Now choose $C_2=2+\max\{x_j\; : \; j\le n_1\}$. Obviously $C_2\ge 2$ and you can choose $n_2$ such that $n_2>n_1$ and $x_{n_2}>C_2$.
You continue by induction. In the $k$-th step you choose $C_k=k+\max\{x_j \; : \; j\le n_k\}$. This means that $C_k\ge k$ and you there exists $n_{k+1}>n_k$ with $x_{n_k}>C_k$.
In this way you obtain a subsequence $n_k$ such that $x_{n_k} \ge k$. This implies that this subsequence converges to $+\infty$.