A sequence {$f_n$} is said to be Cauchy if for every $\epsilon >0$ there is an integer $N$ such that for $n,m \in \Bbb N$ then $|f_n - f_m| < \epsilon$ when $n>m>N$
So we know that $f_J \to f$
Then by the triangle inequality $|f_n - f_m| \le |f_n - f| + |f_m - f|$
Since we know $f_J$ converges to $f$, then by definition $|f_J - f| < \frac{\epsilon }{2}$ where $\frac{\epsilon }{2}$ is arbitrary
Finally, $|f_n - f_m| \le |f_n - f| + |f_m - f| < \frac{\epsilon }{2} + \frac{\epsilon }{2} \to |f_n - f_m| < \epsilon$
This shows that {$f_n$} is Cauchy.
You need to show that for your specific function, $|f_J - f| < \frac{\epsilon }{2}$ can be interpreted to be bounded above by this arbitrary $\frac{\epsilon }{2}$ which will show that it too 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 $d(x,x_n) \le \epsilon$.
Best Answer
If a sequence in this space is Cauchy, then it converges. That is, $\Bbb R^{p \times q}$ under $\|\cdot\|_{1,1}$ (which is isometric to $\Bbb R^{pq}$ under $\|\cdot\|_1$) is indeed a complete metric space.
In general: the finite Cartesian product of complete metric spaces will be complete (this is not true, however, for arbitrary products).
That being said, the condition you provided is insufficient to guarantee that the sequence is Cauchy. As a counterexample, consider the sequence in $\Bbb R^{2 \times 1}$ given by $C_n = (\cos \theta_n,\sin\theta_n)$ where $$ \theta_n = \sum_{i=1}^n \frac 1i $$ Noting that $\|C_n\| < \sqrt{2}$ for each $n$.