A slight infinite extension of this Show that two bounded sequences have convergent subsequences with the same index sequence
Let $S$ be a compact metric space. Suppose, for each $m\in\mathbb{N}$, $\{x(m,n)\}_{n\geq 1}$, is a sequence in $S$. The question is can we always get a strictly increasing sequence of natural numbers $n_k, k=1,2,\cdots$ such that the subsequence for each $m$, $\{x(m,n_k)\}_{k\geq 1}$ is convergent as $k\to \infty$? I want the same $n_k$ working for any $m$.
I worked like this:
Compact implies sequentially compact in metric spaces. Therefore, the sequence for fixed $m$, $\{x(m,n)\}_{n\geq 1}$ has convergent subsequence. But the subsequence indices would differ for each $m$. I am not able to move further inductively. Any help would be grateful.