Prove that a metric space in which every infinite subset has a limit point is separable. 
Definition:

Call the set in the hint E. I've already proven that E is finite and X can be covered by finitely many neighborhoods of radius $\delta$.
What I'm not sure of is the next part, where I n need to show that E is dense. I want to show that every point of X can be a point of E. This happens because for a point x $\in$ X such that x $\in$ $N_\delta(q)$ for some q $\in$ E, we can than choose another smaller $\delta$ so that E also counts x amongst its member.
So for $\delta = 1/n \ (n = 1,2,3,...)$, every point of X can be a point of E. Hence E is dense.
This argument seems to go in the direction of the hint, but I'm not really sure about it. Since wouldn't we need to prove that E is dense for a fixed $\delta$?
 A: The sketched argument is as follows: let $\delta >0$ be given. 
Do the recursion: pick $x_0 \in X$ and having picked $x_0,\ldots,x_n$, pick $x_{n+1}$ such that $$\forall i\le n: d(x_i, x_{n+1}) \ge \delta \tag 1$$
If this process never stops, it's easy to verify that the set $\{x_n: n \ge 0\}$ is infinite and has no limit point.
So for each fixed $\delta>0$ there is a finite stage $n$ where this process breaks down, so we have $x_0,\ldots,x_n$ but for all $x \in X$ we have that $d(x, x_i)< \delta$ for some $i \in \{0,\ldots,n\}$ (or we could have picked $x$ as the $x_{n+1}$ and the process would have gone on). This can be reformulated, as your text does, as 

If $(X,d)$ satisfies the hypothesis that every infinite set has a limit point, then for every $\delta>0$ there are $x_0,\ldots, x_n \in X$ such that $X = \cup \{B(x_i, \delta): i = 0,\ldots, n\}$.

(or in general topology terms: a limit point compact metric space is totally bounded.)
Now apply the above lemma to $\delta=1,\frac12, \frac13,\ldots$ and every time collect the promised centres of the balls as the (finite) subset $F_n$ of $X$ (for $\delta=\frac1n$), and define $D = \cup_n F_n$. The claim now is that $D$ is dense: to see this, pick any $x\in X$ and any $\varepsilon>0$ and we want to show that $B(x,\varepsilon) \cap D \neq \emptyset$: pick $N$ so large that $\frac1N < \varepsilon$ and note that $x \in B(x_i, \frac1N)$ for some $x_i \in F_N$, so that $x_i \in D$ and $d(x,x_i) < \frac1N < \varepsilon$ and so $D \cap B(x,\varepsilon) \neq \emptyset$ is witnessed by this $x_i$.
A: Your argument is nearly valid. It doesn't show that every $x \in X$ belongs to $E$, but rather that every $x$ is in the closure of $E$. And that's ok, because this is the definition of a dense subset: $E$ is dense in X if $\bar{E}=X$.
Also, having said that, you cannot fix $\delta$. That is because of the definition of the closure of $E$: $x\in \bar{E}$ iff $\forall_{\delta >0} N_{\delta}(x)\cap E\neq\emptyset$.
A: Let indeed $x_1$ be arbitrary, and set $\delta_1=1$. Denote $X_1$ the finite set $\{x_1,\dots, x_{n_1}\}$ obtained as in the hint for $\delta_1$. 
Let $x_{n_1+1}, \dots$ be points satisfying $d(x_k, x_j) \ge\delta_2:=\frac12$. Denote their set $X_2$, it's finite again. 
Continue with $\delta_n:=\frac1n$. 
Now, the dense set $E$ is considered to be $E:=\bigcup_nX_n$. 
