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?