Prove) Suppose $x_n\to\infty$. Then every subsequence $x_{n_k}\to\infty$. I don't think this statement is true. What if the sequence $x_n = 1$ for all even numbers but $x_n$ is a strictly monotone sequence when $n$ is odd? Then the sequence $x_n$ is a divergent sequence but still has a convergent subsequence, by making $n_k$ a sequence of even numbers. 
Is my counterexample valid?
 A: Let $(x_{\phi (n)}) $ be a subsequence of $(x_n) $ with $x_n\to +\infty $.
for $A>0$, there exist $N>0$ such that
$$n>N\implies x_n>A $$
but $$\phi (n)\ge n $$ thus
$$n>N\implies \phi (n)>N$$
$$\implies x_{\phi (n)}>A $$
$$\implies \lim_{n\to+\infty}x_{\phi (n)}=+\infty .$$
Your counterexample is not good cause
$\lim x_n $ doesn't exist.
A: There's something subtle here. 
If you define "convergence to infinity" as: For every $A\in \mathbb{R}$ there exists $N_A$ such that if $n > N_A$ then $\lVert x_n \lVert > A$, in that case, the statement that you are trying to refute is valid.
The sequence $x_n = n + (-1)^nn$ does not "converge to infinity" by the definition above even though there are arbitrarily large terms for $n$ sufficiently large. Maybe that is the example you were trying to show, but it doesn't adjust to the premise.
Maybe the confusion comes from the fact that "divergence" and "convergence to infinity" are used interchangeably, which is wrong.
A: How do the even terms $x_n =1$ get larger and larger? For example $x_n \to \infty$ requires that from some $N$ onward (meaning all $n>N$) $x_n$ mus be greater than, say, $2345$. Surely there are even $n$ which are $> N$.
