= Solution
A <sequence> converges to $\ell$ precisely when
$$
\boxed{\forall\varepsilon>0\ \exists N\in\mathbb N\ \forall n\geq N:\quad |x_n-\ell|<\varepsilon.}
$$
For a finite collection of <convergent sequences>, choose $N_j$ for the same tolerance $\varepsilon$ in row $j$, and let $N=\max_{1\leq j\leq k}N_j$. If $n\geq N$, every available $y_n^{(j)}$ lies within $\varepsilon$ of $\ell$, and so does any selected $x_n$. \b[Every such finite rowwise selection converges to $\ell$]. The finiteness is what permits a single maximum of the cutoffs.
For the infinite collection, define
$$
\boxed{y_n^{(j)}=\begin{cases}\ell+(-1)^j,&n=j,\\\ell,&n\ne j.\end{cases}}
$$
For every fixed $j$, this <sequence> is eventually exactly $\ell$, and hence converges to $\ell$. But its diagonal selection is $x_n=y_n^{(n)}=\ell+(-1)^n$, which has two distinct subsequential limits and \b[does not converge]. A moving row index can always choose the exceptional entry before that row reaches its eventual behavior.
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