= Solution
Construct the nested index sequences inductively. Start with $N^{(0)}=(1,2,\ldots)$. Having chosen $N^{(i-1)}$, apply the <Bolzano-Weierstrass theorem> to the bounded sequence of $i$th coordinates along those indices, and let $N^{(i)}$ be an increasing subsequence along which that coordinate converges. Because each new sequence is nested inside all earlier ones, coordinates $1,\ldots,i$ all converge along $N^{(i)}$.
Now take the <diagonal subsequence argument>
$$
m_j=n_j^{(j)}.
$$
Nestedness implies
$$
m_{j+1}=n_{j+1}^{(j+1)}\geq n_{j+1}^{(j)}>n_j^{(j)}=m_j,
$$
so $(m_j)$ is strictly increasing. For every fixed $i$, the tail $(m_j)_{j\geq i}$ is a subsequence of $N^{(i)}$; hence $(x_{m_j}^{(i)})$ converges. This single diagonal subsequence works for every coordinate.
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