I am having some problems with the proof of the following Theorem:
"Let $E$ be a set in a metric space $\mathscr{X}$. Then $E$ has the Lindelöf property provided there exists a countable set $D$ which is dense in $E$ ".
A set $E\in\mathscr{X}$ has the Lindelöf property if every open cover has a countable subcollection that covers $E$. The proof I am using can be found on page 80 of the book "Topological Ideas" by K.G. Binmore. I have a real problem only with one part of the proof, but I show the whole proof as given in the text for completeness (I have made some notational changes). For a proof for $\mathscr{X}=\mathbb{R}^{n}$ see How to prove that if $D$ is countable, then $f(D)$ is either finite or countable?.
Let $\mathscr{U}$ be any collection of open sets that covers $E$. We need to show that a countable subcollection of $\mathscr{U}$ covers $E$. Let $\mathscr{B}$ be the class of open balls $B_{q}(d)$ with centers $d\in D$, and rational radii $q$ such that $B_{q}(d)\subset U$ for at least one $U\in\mathscr{U}$ (I will use $B\in\mathscr{B}$ for a general element). Then $\mathscr{B}$ is countable (I have omitted some stuff which shows this). The following statement is then proven (this is the bit I am having a problem with):
$E\subset\bigcup_{U\in\mathscr{U}}U\subset\bigcup_{B\in\mathscr{B}}B\hspace{200pt}(1)$.
The above is proven in the following way: Let $u\in U$ for any $U\in\mathscr{U}$. Since $U$ is open, it is contained in an open ball $B_{\epsilon}(u)$ such that $B_{\epsilon}(u)\subset U$. Since $E$ is dense in $D$ we have $d(e,D)=0$ for each $e\in E$, and so we can always find a point $d\in D$ such that $d(u,d)<\frac{1}{3}\epsilon$ (my problem is here: I think this only holds if $u\in E$ correct?). We can also choose a rational number $q$ such that $\frac{1}{3}\epsilon So my question is, for the point $u\in U$ in the above paragraph, how can we always find a point $d\in D$ such that $d(u,d)<\frac{1}{3}\epsilon$ if $u\not\in E$? From the hypotheses of the Theorem I do not see why every $u\in U$ has to be in $E$. I now finish the proof as it is given in the book. Proof continued: Let $f:\mathscr{B}\rightarrow\mathscr{U}$ be chosen so that $B\subset f(B)$ for each $B\in\mathscr{B}$, then $f(\mathscr{B})$ is a countable subcollection of $\mathscr{U}$ (see How to prove that if $D$ is countable, then $f(D)$ is either finite or countable? for a proof of this one). Thus $E\subset\bigcup_{B\in\mathscr{B}}B\subset\bigcup_{B\in\mathscr{B}}f(B)\subset\bigcup_{U\in f(\mathscr{B})}U\subset\mathscr{U}$. This shows that $f(\mathscr{B})$ is a countable subcollection of $\mathscr{U}$. I have one question regarding the above paragraph: Can we always choose the function $f$ so that $B\subset f(B)$ for each $B\in\mathscr{B}$? I suppose $f(b)=b$ for all $b\in B$ does the trick? Any help on these questions would be greatly appreciated.