Divergence of subsequence of sequence which diverges I have to prove that given a sequence {an}n, with lim sup an = infinity, prove that there is a sub-sequence ank of an that diverges to infinity.
I used the Theorem which states that for any sequence which has a limit (possibly infinite), all of it's sub-sequences have the same limit. As an clearly diverges to infinity, the limit of all sub-sequences is infinity, and thus all sub-sequences diverge to infinity.
Does this work, or is there something I am shortcutting passed something important?
 A: No, it doesn't work; mainly because the problem statement does not require the sequence to have a limit $+\infty$.
The symbol $\limsup$ doesn't mean a limit, but rather an 'asymptotic upper bound'.
See Wikipedia Limit superior and limit inferior for definitions.
In your case the sequence is said to be unbound from above (it reaches larger and larger values), but not necessarily to converge to $+\infty$. As an example sequences
$$a_n = \begin{cases} n & \text{if }\ 2\vert n\\0&\text{otherwise}\end{cases}$$
or
$$a_n = n\cdot(-1)^n$$
both satisfy $$\limsup_{n\to\infty} a_n = +\infty$$ although none of them has $\lim a_n = \infty$.
Instead, the former one has $\liminf = 0$ whilst the latter has $\liminf = -\infty$.
A: You can't use that theorem, since $\lim_n a_n = +\infty$ isn't necessarily true as CiaPan has pointed out. You can however construct a subsequence as follows: because 
$\limsup_n a_n = \lim_n \sup \{ a_k \mid k \geq n\} = +\infty$, for every $C > 0$ there exists a $N \in \mathbb{N}$ such that $s_m : = \sup \{a_k \mid k \geq m\} \geq C$ for all $m \geq N$. 
So there is some $n_1 \in \mathbb{N}$ such that $s_{n_1} \geq 2$, then there must be some $m_1 \geq n_1$ such that $a_{m_1} \geq 1$ by definition of the supremum. Now take $n_2 > n_1$ such that $s_{n_2} \geq 3$, again there must be some $m_2 \geq n_2$ such that $a_{m_2} \geq 2$. Continuing this process results in a subsequence $(a_{m_k})_k$ such that $a_{m_k} \geq k$ and so we must have $\lim_k a_{m_k} = +\infty$.
