A property of a separable space Suppose $(X,\tau)$ is a separable topological space. Let $C$ be a basis for the topology $\tau$. Is it true that any open set in $\tau$ can be written as a countable union of elements in $C$?
If not, is the above true in case of a separable metrizable space?
Thanks,
Phanindra 
 A: The answer to the first question is no. Let $\mathscr{D}$ be an uncountable family of almost disjoint subsets of $\Bbb N$. Let $X=\Bbb N\cup\mathscr{D}$. Each $n\in\Bbb N$ is isolated in $X$, and a basic open nbhd of $D\in\mathscr{D}$ is any set of the form $\{D\}\cup(D\setminus F)$ such that $F$ is a finite subset of $D$. It’s easy to check that this does define a completely regular topology on $X$. $\Bbb N$ is clearly a countable dense subset of $X$, so $X$ is separable. Let $$\mathscr{B}=\Big\{\{n\}:n\in\Bbb N\Big\}\cup\Big\{\{D\}\cup(D\setminus F):D\in\mathscr{D}\text{ and }F\subseteq D\text{ is finite}\Big\}\;;$$ then $\mathscr{B}$ is a base for $X$, but $X$ itself is not the union of any countable subset of $\mathscr{B}$.
The answer to the second question is yes. Every separable metric space is second countable, i.e., has a countable base. Second countability is clearly a hereditary property, and every second countable space is clearly Lindelöf, so every separable metric space is hereditarily Lindelöf. Now let $\mathscr{B}$ be any base for the separable metric space $X$, and let $U$ be any non-empty open set in $X$; there is a subset $\mathscr{U}\subseteq\mathscr{B}$ such that $U=\bigcup\mathscr{U}$. $\mathscr{U}$ is an open cover of $U$, and $U$ is Lindelöf (as a subspace of the hereditarily Lindelöf space $X$), so $\mathscr{U}$ has a countable subcover $\mathscr{U}_0$. But then $\mathscr{U}_0$ is a countable subfamily of $\mathscr{B}$ whose union is $U$.
Added: One way to get an uncountable almost disjoint family of subsets of $\Bbb N$ is as follows. For each irrational number $x$ let $\langle q_x(k):k\in\Bbb N\rangle$ be a sequence of rational numbers converging monotonically to $x$. It’s easy to see that if $x$ and $y$ are distinct irrationals, the sequences $\langle q_x(k):k\in\Bbb N\rangle$ and $\langle q_y(k):k\in\Bbb N\rangle$ can have only finitely many terms in common. Now let $\varphi:\Bbb Q\to\Bbb N$ be any bijection, for each irrational $x$ let $D(x)=\{\varphi(q_x(k)):k\in\Bbb N\}$, and let $\mathscr{D}=\{D(x):x\in\Bbb R\setminus\Bbb Q\}$; $\mathscr{D}$ is an uncountable family of almost disjoint subsets of $\Bbb N$.
A: I've been trying to come up with a simpler example to answer your first question (Brian's answer is of course complete and great but I still fancied a simpler example) so here it is:
Take $X = \mathbb R$ and the topology $T_B$ with the basis 
$$ B = \{ \{x\} \cup (\mathbb Q \cap (x - \varepsilon, x + \varepsilon)) \Big \vert \text{  where } x \in \mathbb R, \varepsilon > 0\}$$
Then $\mathbb Q$ is dense in $\mathbb R$ in $T_B$ (easy to see) but you cannot write $\mathbb R$ as a countable union of sets in $B$ because each irrational has its own set in $B$ so to cover them all you need at least one set from $B$ per irrational number. 
A: A simple counterexample for your first question is the uncountable particular point topology (example 10 in Steen and Seebach's Counterexamples in Topology).  Let $A$ be an uncountable set, and let $X$ be the disjoint union of $A$ with some additional point $p$.  Define a topology on $X$ by taking as open any set which is either empty or contains $p$.  Then $X$ is separable (indeed, $\{p\}$ is dense in $X$).  Now let $C = \{\{x,p\} : x \in X\}$.  This is a basis for the topology of $X$, but for instance, $X$ itself cannot be written as a countable union of sets from $C$.  (It's worth noting, though, that $X$ is first countable: the single set $\{x,p\}$ is a neighborhood basis at $x$).
For your second question, one can give a proof using simpler words :)  Of course a separable metrizable space is second countable, so has a countable basis $\{U_n\}$.  Let $C$ be any basis and $V$ be open.  For every $x \in V$, we can find $W_x \in C$ with $x \in W_x \subset V$.  We can also find $U_{n(x)}$ with $x \in U_{n(x)} \subset W_x$.  We can now choose a countable set $E$ such that $\{n(x) : x \in E\} = \{n(x) : x \in V\}$.  (More formally, think of the function $n : V \to \mathbb{N}$; for each $k$, let $E$ contain one element of $n^{-1}(k)$ if it is nonempty.)  Now for any $y \in V$, there exists $x \in E$ with $n(x) = n(y)$, so $y \in U_{n(y)} = U_{n(x)} \subset W_x$.  Thus $\bigcup_{x \in E} W_x = V$, so $V$ can be written as a countable union of sets from $C$.
Note that this works in any second countable space, metrizable or not.
A: Let $X=\mathbb{R}_{\ell}$ be the Sorgenfrey line, this is separable but not second countable. It is true if $X$ is metrizable because in metrizable spaces separable and second countable are equivalent.
