Separability is a topological notion. It's usefulness arises from each of the two parts of its definition:
- Countable sets are usually easier to work with than uncountable ones.
- Knowing the behavior of something on a dense set is often useful to prove something about it's behavior on the whole space.
We can think of separability as a way of saying that our topological space doesn't have "too many" points, because every point is just a limit point of some countable subset.
In a metric space, we can actually say more.
A metric space is separable iff it is second countable.
While we think of separability as saying that our space doesn't have "too many" points, we can think of second countability as saying that our space doesn't have "too many" open sets.
One simple result from topology that is very useful comes to mind:
If $f,g:X\to Y$ are continuous functions between topological spaces that agree on a dense subset of $X$ and $Y$ is Hausdorff, then $f=g$.
Thus knowing how a function actions on a dense set will determine it's behavior on the whole space. This type of observation allows for the construction of a metric in each of the cases of the following theorem for normed spaces.
Let $X$ be a normed space. If $X^*$ is separable, then $B_X$ is weakly metrizable. If $X$ is separable, then $B_{X^*}$ is weak* metrizable.
Combined with Banach-Alaoglu's theorem and reflexivity, we obtain results about the compactness of the weak topology. This is the Eberlein–Šmulian theorem.
Separability is also useful in the context of Banach spaces because of the notion of a Schauder basis. Only separable Banach spaces admit a Schauder basis, which allows us to write every element as a unique infinite linear combination of basis elements.
The idea is correct, but not well argumented, I think. It's mostly a problem of notation, however, plus a weakness I'll underline later on.
I'd follow the hint, that is, proving first the space has a countable basis.
For all integers $n>0$, the open cover $\{N_{1/n}(p):p\in K\}$ has a finite subcover; let $X_n=\{x_{n,1}, x_{n,2}, \dots, x_{n,m(n)}\}$ be such that
$$
K=\bigcup_{i=1}^{m(n)}N_{1/n}(x_{n,i})
$$
I claim that the set
$$
\mathcal{B}=\bigcup_{n>0}\bigl\{N_{1/n}(x_{n,i}):1\le i\le m(n)\bigr\}
$$
is a countable basis for $K$. Countability is obvious. Let $p\in K$ and $\varepsilon>0$; we want to prove that there exist $n>0$ and $i$ with $1\le i\le m(n)$ such that $N_{1/n}(x_{n,i})\subseteq N_\varepsilon(p)$.
Take $n$ such that $1/n<\varepsilon/2$. Then $p\in N_{1/n}(x_{n,i})$, for some $1\le i\le m(n)$. If $q\in N_{1/n}(x_{n,i})$, then
$$
d(p,q)\le d(p,x_{n,i})+d(x_{n,i},q)<\frac{1}{n}+\frac{1}{n}<\varepsilon
$$
so $N_{1/n}(x_{n,i})\subseteq N_\varepsilon(p)$ (this is a point where your proof is weak).
Now every metric space having a countable basis is separable. It suffices to take a point in each (nonempty) member of the basis and this is a dense subset, because each open set is the union of members of the basis, so it intersects this countable subset.
Best Answer
Zettili misspoke, I think. Yes, there will be a sequence $(\phi_n)_n$ from $H$ such that for all $\phi \in H$ and all $\varepsilon >0$ we can find some $n$ such that $\|\phi_n - \phi\| < \varepsilon$, which is just a restatement of the general topology concept of separability in the Hilbert space context.
What we cannot have that this sequence will be Cauchy. If it would be it would converge to some $\phi_0 \in H$ and that kills the approximation property. He must have meant something else.
For Hilbert spaces we can also say that separability in an infinite dimensional Hilbert space implies there is an orthonormal basis $(e_n)_n$ which means
The latter form you will probably also meet. It follows from a countable dense set plus the extra structure a Hilbert space has compared to a "mere" topological space.