I need help with the definition of the metric.
For $(x_n)$ in the metric space ${F}(\mathbb{N},\mathbb{R})$ let $$ d(x,y)=\sup\left\{\min\left\{|x_n-y_n|,\frac{1}{n}\right\}\mathrel{\Bigg|} n\in\mathbb{N}\right\} $$ Show that $(x_n)=(x_{n1},x_{n2},\ldots)$ converges to $x=(x_1,x_2,\ldots)$ in the metric space if and only if $\lim_{n\to\infty}x_{nj}=x_j$ for all $j\in\mathbb{N}$.
$\implies$: Per definition we have that $$ \forall\epsilon>0\exists N\in\mathbb{N}\forall n\ge N:d(x_n,x)\le\epsilon $$ Consider some $j\in\mathbb{N}$. Given $\epsilon>0$, $x_n\to x$ is guaranteed by some $N\in\mathbb{N}$. It follows that $d(x_{nj},x_j)\le d(x_n,x)<\epsilon$ for all $n\ge N$.
$\impliedby$: Let there be $k>0$ coordinates in $x_n$. Given $\epsilon > 0$, $\lim_{n\to\infty}x_{nj}=x_j$. Per definition there exists some $N_j$ such that $\forall\dfrac{\epsilon}{k}>0\exists N_j\in\mathbb{N}\forall n\ge N_j:d(x_{nj},x_j)<\epsilon$. Let $N=\max\{N_1,\ldots,N_k\}$. Then for all $n\ge N$ $$ d(x_n,x)\le d(x_{n1},x_1)+\ldots+d(x_{nk},x_k)<k\frac{\epsilon}{k}=\epsilon $$
I do not at any time use the 'special' definition of the metric. Is this somewhat correct? Any corrections are appreciated.
Maybe it would be conceptually easier if you wrote $x_n\to x$ as $n\to\infty$ as the condition $$ \forall m\in\mathbb N, \exists N\in\mathbb N\text{ such that }\forall n\geqslant N, \,\min(|x_{ni}-x_i|,1/i)<1/m\,\forall i\in\mathbb N. $$
$\implies$: for a specific $i$, you need to pick an $m>i$ and then $x_n\to x$ implies that $x_{ni}\to x_i$ in the metric on the real numbers, $d_{\mathbb R}(x,y)=|x-y|$.
$\impliedby$: For a choice of $m$, one needs an $N$ large enough such that the finite sequence of $(x_{n1},\dots,x_{nm})$ has each coordinate no more than $1/m$ distant from $(x_1,\dots,x_m)$. The coordinate convergence and finiteness guarantees this exists.