$\{e_1, e_2, \dots \}$ is closed set in $\ell^2$. Why? The explanation given in the book is that this set of points has no accumulation point. I don't see how this is true. Also, it says that for this reason again the set is not compact and I don't understand this reasoning either. Thanks for your help.
 A: Let's recall the definition of accumulation point:

Given a metric space $(X,d)$ and a subset $A\subseteq X$, an accumulation point of $A$ is a point $p$ of $X$ such that for all $\varepsilon>0$, $(B_{\varepsilon}(x)-\{x\})\cap A$ is non-empty, i.e., there is a point $a_{\varepsilon}\in A$ different from $x$ such that $d(x,a_{\varepsilon})<\varepsilon$. The set of all accumulation points of $A$ is the derived set of $A$, $A'$.

Note that an accumulation point generalizes the notion of closure point as it is a point that is infinitely near to $A$ in a strict sense, meaning that is not infinitely near because the point is in $A$. It is quite easy to show (follows from the definitions) that
$$\overline{A}=A\cup A'\text{,}$$
where $\overline{A}$ is the closure of $A$. So we have:

A set is closed if and only if it contains its accumulation points.

Respect compactness, this follows from the next generalization of Bolzano-Weiertrass' theorem:

Let $(X,d)$ be a metric space and $K$ a compact subset, then any infinite subset $A$ of $K$ has an accumulation point in $K$. In particular, compact infinite sets have accumulation points.

Its proof is by contradiction. As $A'$ is empty, $A$ is closed and so compact as a subset of a compact set. However, as $A'$ is empty, for every $a$ in $A$, we can find an open ball around $a$ $B_a$ such that $B_a\cap A=\{a\}$. This gives a contradiction as then $\{B_a\}_{a\in A}$ is an open cover of $A$ without finite subcoverings.
In your case, just recall that you are working with the metric space $(\ell^2,d)$, where $d(x,y)=||x-y||_2$. Note that the ideas above can be easily applied to your case directly.
