Show that the space of all real sequences with finitely many nonzero is not complete with a specific norm 
Consider the space $\ell_{0}^{2}$ of all real sequences with only a
  finite number of nonzero terms. Show that the space is not complete,
  with the norm $\|x\|={\left(\sum|x|^{2}\right)}^{1/2}$.

My attempt: I've had a break for some time, so I am a bit rusty. But, I really just have to find a Cauchy sequence that does not converge. I was thinking about a sequence where $x_n=1/n$, the first $n$ terms and then zero (it has helped me before, but I'm not sure whether it works). Assume $m>n$, then
$$
d^2(x_m,x_n)=\|x_m-x_n\|^2=\sum_{j=n}^m\frac{1}{j^2}\leq\frac{m-n+1}{n^2}
$$
which we can make arbitrarily small, and therefore it is a Cauchy sequence. (right?)
So, I'm thinking that obviously the limit doesn't exist, since it would not have finitely many zeros, and therefore it does not converge. Is my reasoning OK? If so, how do I continue?
Alternative approach: If you have some alternative approach I'm happy to hear about it.
 A: I think your approach is fine. One way to finish is to observe that the space $l^{2}_{0}$ is contained in $l^2$. Your limit sequence is in $l^2$ but not $l^{2}_{0}.$ Since a sequence has a unique limit in $l^2$ it can't have a limit in $l^{2}_{0}$ so $l^{2}_{0}$ is not complete.
A: Yes. Take any sequence $(a_j)_j$ of positive reals such that $\sum_{j\in \Bbb N}a_j^2<\infty.$ Let $x_n=(x_{n,j})_j$ where $x_{n,j}=0$ if $j>n$ and $x_{n,j}=a_j$ if $j\le n.$ Then $$\lim_{n\to \infty}\sup_{m>n}\|x_n-x_m\|^2=\lim_{n\to \infty}\sum_{j=n+1}^{\infty}a_j^2=0.$$ So $(x_n)_n$ is a Cauchy sequence.
If $y=(y_j)_j\in l^2_0$ and $\lim_{n\to \infty}\|x_n-y\|=0,$  then for each $j,$ we have $\lim_{n\to \infty}|x_{n,j}-y_j|=0.$
Otherwise, for at least one $j$, there would be some $r>0$ such that $|x_{n,j}-y_j|>r$ for infinitely many $n.$  But then $\|x_n-y\|\ge |x_{n,j}-y_j|\ge r$ for infinitely many $n.$ 
So if $y=\lim_{n\to \infty}x_n$ exists in $l^2_0$ then $y_j=a_j\ne 0$ for all $j,$ so $y\not \in l^2_0,$ a contradiction. 
Remark: Set-theoretically, a sequence is a function $f$ with domain $\Bbb N$ (or some other suitable countably infinite domain). So let $x_n=f_n$ where $f_n(j)=x_{n,j}$ and let $y_j=g(j).$ If $(f_n)_n$ converges in norm to $g$ then $f_n$ converges point-wise to $g.$ That is, $f_n(j)\to g(j)$ for each $j,$  Because $|f_n(j)-g(j)|=|x_{n,j}-y_j|\le  \|x_n-y\|\to 0 $ as $n\to \infty.$
