 Solution

Interpret limit points as subsequential limits, including a value repeated infinitely often. Let $L$ be the common nonempty compact cluster set. Boundedness implies $\mathrm{dist}(a_n,L)\to 0$ and likewise for $b_n$: otherwise a subsequence staying a fixed distance away would have a convergent further subsequence whose limit both belongs to $L$ and stays away from it.

Construct a matching of indices in alternating steps. At step $2j-1$ choose the least unused index $m$ in the first sequence. Choose a nearest point $\ell \in L$ to $a_m$ and an unused index $n$ in the second sequence with $|b_n-\ell |<1/j$. Such indices exist infinitely often because $\ell$ is a subsequential limit. Pair $m$ with $n$; their values differ by at most $\mathrm{dist}(a_m,L)+1/j$. At step $2j$ choose the least unused second-sequence index and match it to an unused first-sequence index within $1/j$ of a nearest point of $L$. This gives the analogous bound with $b_n$.

Every index is used exactly once: no step reuses an index, and by step $2j$ at least the first $j$ indices in each sequence have been used. Hence the produced lists are permutations satisfyingThe indices actively chosen as least unused tend to infinity on their respective odd and even steps. Their distances to $L$ therefore tend to zero, and $1/j\to 0$. The two matching estimates prove $\boxed{|a_{m_k}-b_{n_k}|\to 0}$. The cluster-set convention is important here; accumulation points of the set of distinct values would ignore infinite repetitions.