# Existence of common convergent subsequence indexes for infinite number of sequences in a compact metric space.

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.