TITLE: Second-countable implies separable/Axiom countable choice
QUESTION [14 upvotes]: Let $(X,\mathscr T)$ be a topological space, and $(B_n)_{n\ge1}$ a
  countable basis for X. Under this assumptions, X is separable.

The proof of this assertion is as follows:
We can assume without loss of generality that all the $B_n$ are nonempty, because the empty ones can be discarded. Now, for each $B_n$, pick any element $x_n \in B_n$. Let $D$ be the set of these $x_n$. $D$ is clearly countable. We claim that $D$ is dense in $X$.
To see this, let $U$ be any nonempty open subset of $X$. Then, $U$ contains some $B_n$, and hence, $x_n \in U$. But by construction, $x_n \in D$, so $D$ intersects $U$, proving that $D$ is dense. $\blacksquare$
My question is, can this theorem be proven without the axiom of countable choice?

REPLY [13 votes]: No. It cannot be proved without the axiom of choice that every second countable space is separable. In fact the following are equivalent:

The axiom of countable choice.
Every second countable space is separable.

For a related topic (with references), Does proving (second countable) $\Rightarrow$ (Lindelöf) require the axiom of choice?
Or the following paper:

Horst Herrlich, Choice principles in elementary topology and analysis
Comment. Math. Univ. Carolin 38,3 (1997) 545-552.


It is consistent (with the failure of choice) that there is a subset of the real numbers which is infinite Dedekind-finite, that is not finite and does not have any countable infinite subset.
Take $D$ be to such subset, then it is easy to show that $D$ in the relative topology is second-countable, but it clearly not separable.