closed bounded equivalent to compactness? 
In preparation for an exam I am struggling with the following problem. We let $A:=\{x=(x_{n})_{n}\in \ell^{2}| \phantom{x} \|x\|\leq 1\}$ and consider $d:A\times A\rightarrow \mathbb{R}_{+}$ defined by
$$d(x,y)=\sum_{n=1}^{\infty}(1/2)^{n}|x_{n}-y_{n}|.$$
Prove that $(A,d)$ is sequentially compact.

My question is can I do this by showing it is closed and bounded or is it not allowed because of dimensionality issues? Or is there another way I can show it is sequentially compact without resorting to a diagonal argument?
 A: Let $x = (x_n)_{n \in \mathbb N} \in \ell^2$ denote an element of $\ell^2$ and $x^k = (x_n^k)_{n \in \mathbb N}$ denote a sequence of elements in $\ell^2$. Since for every $x^k$, $\sum_{n \ge 1} |x_n^k|^2 \le 1$, we have $x_n^k \in [-1,1]$ for every element of the sequence. Note that $[-1,1]$ is closed and bounded, hence sequentially compact.
From the sequence $x^k$ satisfying $\| x^k \| \le 1$, the sequence $\{x_1^k\}_{k \in \mathbb N}$ is a sequence in $[-1,1]$, hence admits a convergent subsequence to $x_1$. Choose the sequence such that $|x_1^k - x_1| \le 2^{-k}$. 
Inductively, if you have a subsequence of $x^k$ such that the first $N$ component sequences $\{x_1^k\}_{k \in \mathbb N}$, $\dots$, $\{x_N^k\}_{k \in \mathbb N}$ converge respectively to $x_1, \dots, x_N$ and $|x_i^k - x_i| \le 2^{-k}$ for $k \ge i$, $1 \le i \le N$, keep the first $N$ terms of the subsequence $x^k$ previously chosen as they are, and choose a subsequence that begins at the $N+1^{\text{th}}$ such that $x_{N+1}^k$ converges to $x_{N+1}$ and $|x_{N+1}^k - x_{N+1}| \le 2^{-k}$ for $k \ge N+1$. 
It's important that you require $k \ge i$ and not $k \ge 1$ ; otherwise you cannot "let the algorithm run to infinity". Letting $k \ge i$ allows you to keep the first $i$ terms of your subsequence satisfying the constraints on their components, and at the next iteration of the process, the first $i$ terms can be preserved, and all you need to choose the rest of the sequence, starting at the $(i+1)^{\text{th}}$ term. Also note that if $|x_n^k - x_n| \le 2^{-k}$, then a subsequence of $x^k$, say $x^{k_{\ell}}$, will satisfy $|x_n^{k_{\ell}} - x_n| \le 2^{-k_{\ell}} \le 2^{-k}$ because $k_{\ell} \ge k$. 
In other words, each iteration only builds the $n^{\text{th}}$ term of the subsequence, but in such a way that some bounds on the convergence are obtained along the way.
This gives you a subsequence of $x^k$ and an element of $\ell^2$, namely $x = (x_1,\dots,x_n,\dots)$ (although we still have to prove $\|x\| \le 1$) such that
$$
|x_n^k - x_n| \le 2^{-k}, \quad k,n \in \mathbb N, \quad k \ge n.
$$
Now
$$
d(x^k,x) = \sum_{n \ge 1} \frac{|x_n^k - x_n|}{2^n} = \sum_{n=1}^k \frac{|x_n^k - x_n|}{2^n} + \sum_{n > k} \frac{|x_n^k - x_n|}{2^n} \le \sum_{n=1}^k \frac{1/2^k}{2^n} + \sum_{n > k} \frac 1{2^n}
$$
and
$$
\sum_{n=1}^k \frac{1/2^k}{2^n} + \sum_{n > k} \frac 1{2^n} = \frac 1{2^k} \left( \sum_{n=1}^k \frac 1{2^n} \right) + \sum_{n > k} \frac 1{2^n} \le \frac 1{2^k}(1) + \frac 1{2^k} = \frac 1{2^{k-1}} \underset{ k \to \infty}{\longrightarrow} 0.
$$
Since $\ell^2$ is complete, $x \in \ell^2$ and since $\|x^k\| \le 1$ and $x^k \to x$ in $\ell^2$, $\|x\| \le 1$. 
Hope that helps,
