problem metric involving sequences 
UPDATED: In preparation for an exam I am struggling with the following problem. We have $A:=\{x=(x_{n})_{n}\in \ell^{2}| \phantom{x} \|x\|\leq 1\}$. Consider the metric $d:A\times A \rightarrow \mathbb{R}_{+}$ defined by 
  $$d(x,y)=\sum_{n=1}^{\infty}(1/3)^{n}|x_{n}-y_{n}|.$$
  Prove that $(A,d)$ is sequentially compact.

What I have done so far:
I have updated what I have done so far using the hints below. But I still have some questions as I want to make sure I understand it thorougly.
Let $(x_{n})_{n}$ be an arbitrarily chosen sequence in $A$. Now first take a subsequence $(x_{n1})_{n}$ such that the first component converges. Now take a subsequence of this subsequence $(x_{n2})_{n}$ such that the second component converges. 
Going on in the same way we get a sequence of nested subsequences $(x_{n(k+1)})_{n}\subset (x_{nk})_{n}$. Now take the sequence $(x_{kk})_{k}$, this construction guarantees that every component of this sequence converges, it is equivalent to the convergence in the metric. Also note that for each $k,n$ that $x_{nk}$ is a sequence of real numbers so we have in fact three layers of sequences. Since the elements are sequences and we take sequences of these elements and than a sequence of subsequences of the sequence of elements in $A$. Hence, $(A,d)$ is sequentially compact.
(Question: Have I now formally proved that indeed $(A,d)$ is sequentially compact)
I don't have much experience with these kind of proofs so any help is much appreciated.
 A: (Note: The question was substantially altered since I wrote this answer. If you wish to see what I really answered, use the history mechanism to see the original question.)
You got it wrong from the start. Let's drop the comma and just write $x_n=(x_{nj})_j$, shall we? Now you want convergence in the first component, so you'd like to have $x_{n1}$ convergent as $n\to\infty$. It won't be convergent necessarily, but you can pick a subsequence that converges. Note that this really means picking a subsequence of $(x_n)$ so that $(x_{n1})$ converges. Next, pick a subsequence of the subsequence so that $(x_{n2})$ converges, and so on. Finally, “take the diagonal” to finish the proof.
Incidentally, I think the conventional notation for subsequences is a horrible mess, and makes this kind of proof much harder than necessary, both to write and to read. So I have written up an alternative view of the diagonal method. The basic idea: All the difficulties stem from the insistence of relabeling indices when taking a subsequence, so that the sequence is always indexed on the set of natural numbers. By allowing subsets of the natural numbers as index sets, we simplify this tremendeously.
A: You need to show something that's bounded with respect to the $d$ metric has a convergent subsequence, not that something that's bounded with respect to the normal $l^2$ norm has a convergent subsequence. 
