Separable iff Lindelöf for pseudometric spaces I'm trying to prove, for $X$ a pseudometric space
$$X \text{ Lindelöf } \Leftrightarrow X \text{ separable }$$
Here are some of my ideas so far - the forward direction should work:
$(\Rightarrow)$ For each $n \in \mathbb{N}^+$ the cover of balls $\lbrace B(x,\frac{1}{n}) \mid x\in X \rbrace $ has a countable subcover $\lbrace B(x^n_i , \frac{1}{n}) \mid x^n_i \in X , i \in \mathbb{N} \rbrace $. Then for every $n \in \mathbb{N}^+$, $i \in \mathbb{N}$ choose $d_{n,i} \in B(x^n_i , \frac{1}{n})$. It should not be hard to show that $D = \lbrace d_{n,i} \mid n \in \mathbb{N}^+ , i \in \mathbb{N}\rbrace$ is dense in $X$. Any suggestions to improve this notation? I feel like this notation is unnecessarily ugly. 
$(\Leftarrow)$  Let $D$ be a countable dense subset, and $(U_i)_{i\in I}$ a cover of $X$. Now I'm not so sure what to do. We know we can cover with countably many balls centred at points in $D$, but how to connect this with the cover $(U_i)_{i\in I}$ ?  
 A: For the forward direction you can get rid of some of the notational clutter as follows.

For $n\in\Bbb Z^+$ the set of open $\frac1n$-balls is an open cover of $X$. Since $X$ is Lindelöf, for each $n\in\Bbb Z^+$ there is a countable $D_n\subseteq X$ such that $\left\{B\left(x,\frac1n\right):x\in D_n\right\}$ covers $X$. Let $D=\bigcup_{n\in\Bbb Z^+}D_n$; $D$ is the union of countably many countable sets, so $D$ is countable. To see that $D$ is dense in $X$, let $U$ be any non-empty open set in $X$, and fix $y\in U$. There is an $n\in\Bbb Z^+$ such that $B\left(y,\frac1n\right)\subseteq U$. By the definition of $D_n$ there is an $x\in D_n$ such that $y\in B\left(x,\frac1n\right)$. By symmetry $x\in B\left(y,\frac1n\right)\subseteq U$, so $x\in U\cap D$, and $D$ is indeed dense in $X$.

The point is that you don’t actually have to keep track of the individual points of $D$ and where they came from.
For the other direction, I suggest proceeding as you would if $d$ were a metric: given the countable dense set $D$, show that
$$\mathscr{B}=\left\{B\left(x,\frac1n\right):x\in D\text{ and }n\in\Bbb Z^+\right\}$$
is a base for the topology on $X$. You will then have shown that $X$ is second countable and hence (by an easy argument) Lindelöf.
