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?
Let $(x_{\phi (n)}) $ be a subsequence of $(x_n) $ with $x_n\to +\infty $.

for $A&gt;0$, there exist $N&gt;0$ such that

$$n&gt;N\implies x_n&gt;A $$

but $$\phi (n)\ge n $$ thus

$$n&gt;N\implies \phi (n)&gt;N$$
$$\implies x_{\phi (n)}&gt;A $$

$$\implies \lim_{n\to+\infty}x_{\phi (n)}=+\infty .$$

Your counterexample is not good cause
$\lim x_n $ doesn't exist.
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 $n&gt;N$) $x_n$ mus be greater than, say, $2345$. Surely there are even $n$ which are $&gt; N$.