How to show that a Cauchy sequence of sequences converges

There is a notational issue here. You need to show that a Cauchy sequence of elements of $c_0$ converges to an element of $c_0$. It might help to refer to an element of $c_0$ as $x$, and then address a specific element of the sequence as $x(k)$.

So suppose you have $x_n \in c_0$ such that $x_n$ is Cauchy with the distance given above.

For each $k$, you have $|x_n(k)-x_m(k)| \le d(x_n,x_m)$, so there is some $x(k)$ such that $x_n(k) \to x(k)$. This is the candidate sequence.

Now you must show that $x \in c_0$ and $d(x,x_n) \to 0$.

Let $\epsilon>0$, choose $N$ such that if $m,n \ge N$ then $d(x_n,x_m) < { 1\over 3} \epsilon$. Now choose $K$ such that if $k \ge K$, then $|x_N(k) | < { 1\over 3} \epsilon$. Then, for $k \ge K$ \begin{eqnarray} |x(k)| &\le& |x(k)-x_m(k)| + |x_m(k)-x_N(k)| + |x_N(k)| \\ &<& |x(k)-x_m(k)| + {2 \over 3}\epsilon \end{eqnarray} Now choose $m \ge N$ (which will, in general, depend on $k$) such that $|x(k)-x_m(k)| < {1\over 3} \epsilon$, and we see that $|x(k)| <\epsilon$. Hence $x (k) \to 0$ and so $x \in c_0$.

Showing that $x_n \to x$ is similar: Let $\epsilon>0$ and choose $N$ such that if $m,n \ge N$ then $d(x_n,x_m) < { 1\over 2} \epsilon$. Then, for $m,n \ge N$ we have \begin{eqnarray} |x(k)-x_n(k)| &\le& |x(k)-x_m(k)| + |x_m(k)-x_n(k)| \\ &<& |x(k)-x_m(k)| + {1 \over 2} \epsilon \end{eqnarray} Now choose $m \ge N$ (which will, in general, depend on $k$) such that $|x(k)-x_m(k)| < {1 \over 2} \epsilon$, then we see that $|x(k)-x_n(k)| < \epsilon$ for all $n \ge N$, and so $d(x,x_n) \le \epsilon$.

Let $\{x_n\}$ be a Cauchy sequence. Then, $d(x_n,x_m) \to 0$ as $n,m \to \infty$.

Let $c \in \mathbb{N}$. Note that for all $m,n$, $|x_m(c)- x_n(c)| \leq d(x_m-x_n)$.

Hence, $x_m(k)$ is a Cauchy sequence in $\mathbb{R}$, for all fixed $k \in \mathbb{N}$. Moreover, $x_m(k)$ is uniformly convergent, which is to say, that given $\epsilon > 0$, there is $M$ large enough so that $|x_m(k)-x(k)| < \epsilon$ for $m > M$, for all $k \in \mathbb{N}$.

Now, $\mathbb R$ is complete, so there exists $l_k$ such that $x_m(k) \to l_k$.

Define a new sequence $x(k) = l_k$.

See that $d(x_n , x) = \sup |x(d)-x_n(d)|$. By uniform convergence property, this converges to zero, hence $x_n \to x$.

To see that $x \in c_0$, we want that $x(d) \to 0$ as $d \to \infty$. We know that $x(d) = \lim x_n(d)$, so use this knowledge, along with the fact that $x_n(d) \to 0$ as $n \to \infty$, to show that $x(d) \to 0$. (It's a limit switching argument).