Consider the sequence: $x_n=\begin{cases} n& n=2k\\\frac1n &n=2k+1\end{cases}$ this sequence is unbounded, but has a convergent subsequence, and note that $u_n=\frac1{x_n}$ is also unbounded but has a convergent subsequence. Neither $x_n$ nor $u_n$ are convergent themselves.
What can be said true about $(3)$ is that if $x_n$ is unbounded then $\frac1{x_n}$ has a convergent subsequence, simply take $x_{n_k}$ to be a strictly increasing subsequence which is unbounded, and show that $\frac1{x_{n_k}}$ is a bounded sequence.
(And as you said, $(1)$ and $(4)$ are indeed false.)
Here's a more formal (and shorter) proof:
Let $(x_{n_k})$ be a convergent subsequence of $(x_n)$, which we know exists, and let $L$ be its limit. Then, given any $\varepsilon > 0$, there is some $K \in \mathbb{N}$ such that for all $k \geq K$, $|x_{n_k} - L| < \varepsilon$ (this is the definition of the limit).
Now, for any $n \geq n_K$, we will show than $|x_n - L| < \varepsilon$, which will complete the proof. For this purpose, suppose that this is not true. Then there is some $n \geq n_K$ such that $|x_n - L| \geq \varepsilon > |x_{n_K} - L|$.
First, note that if $x_1 \leq L$ then we must have $x_i \leq L$ for all $i$: if not, then we have some $i$, such that $x_1 \leq L < x_i$, but then for all $m > i$, $x_m \geq x_i > L$, so $|x_m - L| \geq |x_i - L| > 0$, so in particular, $(x_{n_k})\not\to L$, a contradiction. Symmetrically, if $x_1 \geq L$, then $x_i \geq L$ for all $i$.
Now, we have a problem: we have either $x_n \geq L + \varepsilon > x_{n_K} \geq L$ or $x_n\leq L - \varepsilon < x_{n_K} \leq L$, but then monotonicity gives us either $x_m \geq L + \varepsilon$ or $x_m \leq L - \varepsilon$ for all $m \geq n$, in particular for all but finitely many $n_k$, so $|x_{n_k} - L| \geq \varepsilon$ for all but finitely many $n_k$, so $(x_{n_k})\not\to L$, a contradiction.
Thus, we must have $|x_n - L| <\varepsilon$ for all $n \geq n_K$, hence $(x_n)\to L$.
Best Answer
In the comments you considered the example $x_n = 0.001 n$ and that correctly eliminated choices A, C, D. So B is the only remaining choice.
It is useful to know why B holds true, that is, why any sequence $\{x_n\}$ that satisfies $x_n/n\rightarrow 0.001$ must be unbounded. Knowing that it is true for a particular example is not good enough. One way to explore this is through "proof by contradiction": Assume $\{x_n\}_{n=1}^{\infty}$ is bounded, so that there is a finite number $M$ such that $$ -M \leq x_n \leq M \quad \forall n \in \{1, 2, 3, ...\}$$ Now reach a contradiction with the assumption $x_n/n\rightarrow 0.001$.