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
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 $\Bbb N$.
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.