Second-countable implies separable/Axiom countable choice
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.
There is an immediate reversal. Let $(A_n)$ be any countable sequence of nonempty sets. For the purposes of countable choice we may assume the sets are pairwise disjoint. Let $T$ be a space whose points are $\bigcup_n A_n$ and whose topology is generated by the basis $\{A_n : n \in \omega\}$. Let $\{ c_m : m \in \omega\}$ be an enumerated countable dense subset of $T$. For each $n$ let $j(n)$ be minimal such that $c_{j(n)} \in A_n$. Then $\{c_{j(n)} : n \in \omega\}$ is a choice set for the sequence $(A_n)$.