Subset of $\ell^2$ which is closed and bounded, but not compact [closed]

Consider the space $\ell^2=\left\lbrace x=(x_n)_{n\in \mathbb{N}}; \sum_{n=1}^{\infty}x_n^2 < \infty\right\rbrace$ with the inner product $$\langle x,y \rangle = \sum_{n=1}^{\infty}x_n\cdot y_n$$ where $x=(x_n)_{n\in \mathbb{N}}$ and $y=(y_n)_{n\in \mathbb{N}}$. Prove that there exists $A\subset \ell^2$ such that $A$ is closed and bounded, but not compact.

I know it's not cool to post questions with at least showing what I've tried. But, all the subsets that I can think of as they are closed and bounded end up being compact. I accept any suggestion or hint.

Solution 1:

In a normed vector space $X$, the unit sphere is compact iff $X$ is finite dimensional.

Solution 2:

How about $X_k=(x_n)$ where $x_k=1$ and $x_n=0$ for $n\ne k$? It is closed, since there are no limit points, bounded since all $||X_k||=1$, but it is not compact, since each point can be overed by an open set and the open sets can all be disjoint.