Equivalence of three properties of a metric space. Another question about the convergence notes by Dr. Pete Clark:
http://alpha.math.uga.edu/~pete/convergence.pdf
(I'm almost at the filters chapter!  Getting very excited now!)
On page 15, Proposition 4.6 states that for the following three properties of a topological space $X$,
$(i)$ $X$ has a countable base.
$(ii)$ $X$ is separable.
$(iii)$ $X$ is Lindelof (every open cover admits a countable subcover).
we always have $(i)\Rightarrow (ii)$ and $(i)\Rightarrow (iii)$.
Also, we if $X$ is metrizable, we have $(iii)\Rightarrow (i)$, and thus all three are equivalent.
This last part confuses me.  We establish all the implications claimed in the proof, but there seems to be a missing link in the claim that all three are equivalent: namely $(ii)\Rightarrow (iii)$.
 A: Note that a separable metric space has a countable basis.  Specifically, we take a countable dense subset $S$ and take the set of balls centered at $s$ with radius $1/n$ for each $n \in N$, $s \in S$.  This can be checked to be a basis.  So then $(ii) \Rightarrow (i)$ is proven, which is the missing link.
A: The missing implication "separable metrizable implies second countable" is rather easy to prove -- as Carl's answer shows -- but the proof should still appear in the notes.  
I have uploaded a new version taking this into account.  Thanks for bringing this to my attention.
A: Carl’s approach is almost certainly the easiest way to patch the gap in the notes. A nice variant is to let $D=\{x_n:n\in\Bbb N\}$ be a countable dense subset of $X$, define a map $$h:X\to\Bbb R^{\Bbb N}:x\mapsto\langle d(x,x_n):n\in\Bbb N\rangle\;,$$ and prove that $h$ is s homeomorphism of $X$ onto a subspace of $\Bbb R^{\Bbb N}$: being a countable product of second countable spaces, $\Bbb R^{\Bbb N}$ is easily shown to be second countable.
It is possible to give direct proofs of $(ii)\implies(iii)$ and $(iii)\implies(ii)$, but every one that I can think of either (a) smuggles in what amounts to a proof of second countability along the way or (b) uses much higher-powered machinery. 
As an example of (a), consider the following proof that $(iii)\implies(ii)$.

For each $n\in\Bbb N$ let $\mathscr{U}_n=\{B(x,2^{-n}):x\in X\}$, the set of open $2^{-n}$-balls in $X$; $\mathscr{U}_n$ is an open cover of $X$, so it has a countable subcover $\mathscr{V}_n=\{B(x_n(k),2^{-n}):k\in\Bbb N\}$. Let $$D=\{x_n(k):n,k\in\Bbb N\}\;;$$ clearly $D$ is countable. Let $W$ be any non-empty open set in $X$. Pick $x\in W$; there is an $n\in\Bbb N$ such that $B(x,2^{-n})\subseteq W$. $\mathscr{V}_n$ covers $X$, so there is some $k\in\Bbb N$ such that $x\in B(x_n(k),2^{-n})$; but then $x_n(k)\in D\cap B(x,2^{-n})\subseteq D\cap W$, and $D$ is dense in $X$. $\dashv$

With just a little more work this becomes the argument that Carl suggested to prove that $(iii)$ implies $(i)$.
As an example of (b), the fact that every metric space has a $\sigma$-discrete base almost immediately implies that every separable metric space is Lindelöf: 

Let $\mathscr{B}=\bigcup\{\mathscr{B}_n:n\in\Bbb N\}$ be a base for $X$ such that each $\mathscr{B}_n$ is discrete. Suppose that $\mathscr{U}$ is an open cover of $X$ with no countable subcover. Let $\mathscr{R}\subseteq\mathscr{B}$ be a refinement of $\mathscr{U}$ covering $X$. $\mathscr{R}$ has no countable subcover, so $\mathscr{R}\cap\mathscr{B}_n$ is uncountable for some $n\in\Bbb N$. But then $\mathscr{R}\cap\mathscr{B}_n$ is an uncountable family of pairwise disjoint, non-empty open sets, and $X$ cannot be separable. $\dashv$

The reason for the difficulty in going directly between $(ii)$ and $(iii)$ is that in general separability and Lindelöfness are very far from being equivalent; in the metric setting it’s second countability that ties them together by being equivalent to each. Here are a couple of examples illustrating their independence, even in rather nice spaces.
If you retopologize $\Bbb R$ by making every $x\in\Bbb R\setminus\{0\}$ and giving $0$ a base of nbhds of the form $A\setminus C$, where $C$ is any countable subset of $\Bbb R\setminus\{0\}$, then $X$ is a very nice space that is Lindelöf but not separable.
On the other hand, a slightly more complicated retopologization of $\Bbb R$ yields a very nice space that is separable but not Lindelöf. Make each rational an isolated point. To each irrational $x$ associate a sequence $\langle q_x(k):k\in\Bbb N\rangle$ of rational numbers converging monotonically to $x$ in the usual topology; a base of the topology at $x$ consists of the sets $B_n(x)=\{x\}\cup\{q_x(k):k\ge n\}$ for $n\in\Bbb N$. Note that if $x$ and $y$ are distinct irrationals, the sets $\{q_x(k):k\in\Bbb N\}$ and $\{q_y(k):k\in\Bbb N\}$ have at most finitely many points in common; this ensures that the space is completely regular and Hausdorff. $\Bbb Q$ is a countable dense subset, so the space is separable. And $$\{B_0(x):x\in\Bbb R\setminus\Bbb Q\}\cup\Big\{\{q\}:q\in\Bbb Q\Big\}$$ is an open cover with no countable subcover, so the space is not Lindelöf.
