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?
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 $(A_n)$.