Suppose that $(u_n)$ is unbounded above. Then, we pick any arbitrary monotone increasing subsequence $(v_n)$ of $(u_n)$. But by hypothesis, we can find a subsequence of $(v_n)$ that converges to $0$. Hence, $\lim v_n =0$. Therefore, $(u_n)$ must be bounded above.
Again, let $(u_n)$ be unbounded below. Then we pick any monotone decreasing subsequence $(w_n)$. By similar arguments, $(u_n)$ must be bounded below.
Being bounded, it has a finite $\limsup =l$ and $\liminf =m$.
Hence, there must be convergent subsequences $(a_n)$ and $(b_n)$ that converge to $l$ and $m$ respectively. But every subsequences of the above two must converge to $l$ and $m$ respectively (treating the two as convergent sequences). Again, by hypothesis, they both have subsequences that converge to $0$. Therefore, both $(a_n)$ and $(b_n)$ converge to $0$.
In conclusion, $\limsup u_n = \liminf u_n = 0 =\lim u_n$
Is this alright?
Best Answer
Here's a more direct approach that also generalizes to limits in any metric space (or any sequential topological space).
Suppose to the contrary that $u_n$ does not converge to zero. Then there is a neighborhood $V$ of $0$ such that infinitely many of the $u_n$ are outside $V$. Consider these infinitely many $u_n$ as a subsequence of the original sequence, $u_{n_1}, u_{n_2}, \dots$. But by assumption, the subsequence $u_{n_k}$ has a further subsequence that converges to 0, which means that infinitely many of the $u_{n_k}$ are in $V$. This is a contradiction.