Please some help with sequences I am thinking about following: 
Can I find an example of sequence that has a bounded subsequence but no convergent subsequence
Then, my first thought is that it would be same if I try to find a bounded sequence that has no convergent subsequence. I failed to write down an example. But: if the sequence is all rational numbers in $[-1,1]$ in a random order, would it not be an example? When it is random it would not converge, or is it wrong to think so? Otherwise is it possible to write down an example without using random order?
 A: It is a property of the reals that any bounded sequence of reals has a Cauchy subsequence.  However, as drhab points out, $\mathbb{Q}$ is not complete. If we consider sequences in $\mathbb{Q}$, the sequence $a_0=1$ and $a_{n+1}=\frac{a_n^2+2}{2a_n}$ is bounded, but not convergent in $\mathbb{Q}$ (the limit in $\mathbb{R}$ is $\sqrt2\not\in\mathbb{Q}$).
A: Take a sequence $\left(a_{n}\right)$ in $\mathbb{R}$ with $a_{n}\in\mathbb{Q}$
for each $n$, and converging to some $a$ with $a\notin\mathbb{Q}$.
Then looking at $\left(a_{n}\right)$ as a sequence in $\mathbb{Q}$
you have a bounded sequence that has no convergent subsequence.
A: Its a theorem that every bounded sequence in $\mathbb{R}^k$ has a convergent subsequence [Rudin Walter, Principles of Mathematical analaysis], so it provides that if you have a sequence with a bounded subsequence then its subsequence has a convergent subsequence(which is a subsequence of the original sequence). Thus the answer of your question in $\mathbb{R}^k$ is ''NO''. And if you want such a sequence you must search in other fields. 
For example consider subspace $(0,1)$ of $\mathbb{R}$, then $\{\frac{1}{n}\}$ is bounded but not convergent in $(0,1)$. Because $0\not \in (0,1)$
