If a sequence is unbounded from above, it has a subsequence with $x_{n_k} > k$ 
Let $(x_n)_{n \in \mathbb{N}}$ be an unbounded sequence in $\mathbb{R}$. If $(x_n)_{n \in \mathbb{N}}$ is unbounded above, show that $(x_n)_{n \in \mathbb{N}}$ has a subsequence $(x_{n_k})_{k\in \mathbb{N}}$ with $x_{n_k} > k$ for all $k$ in $\mathbb{N}$.

I do not know how to start at this problem. Should I start by using the definition of a subsequence diverging to $+\infty$ (for all $n \in \mathbb{N}$ , there exists $M \in \mathbb{R}$ such that $x_n > M$)? After all, the sequence $(x_n)_{n \in \mathbb{N}}$ is stated to be "unbounded above".
Should I use the fact that if $(x_{n_k})_{k \in \mathbb{N}}$ is a subsequence of $(x_n)_{n \in \mathbb{N}}$, then $n_k > k$ for every $n,k \in \mathbb{N}$.

A sequence $(x_n)$ is "bounded above" if there is a number $M$ such that $x_n\le M$ for all $n$. If the sequence is "unbounded above", that means there is no such $M$; in other words, for any number $M$, it is not true that $x_n\le M$ for all $n$; in other words, for any number $M$, there is some $n$ such that $x_n\gt M$.
