To provide some context, I was trying to prove $(X, d)$ separable $\implies$ X second countable, specifically without using any form of choice.
Let $U \subseteq X$ open and $D$ our countable dense subset, and consider $\mathscr B := \{B_q(d) \mid q\in \mathbb Q_{\geq 0}, d\in D\}$, which should be our basis.
In particular, I will verify that $$ U = \bigcup \underbrace{\{B_q(d) \mid q\in \mathbb Q_{\geq 0}, d\in D, B_q(d)\subseteq U\}}_{\mathscr S}. $$
Proving $\supseteq$ is trivial. I now proceeded to prove that for every $u\in U$, I can find some ball $B_\delta(u)\subseteq U$. To avoid any choice, set $\delta := \sup\{\delta^\prime\mid B_{\delta^\prime}(u)\subseteq U\}$. However, now we have to choose a point $x\in D\cup B_\delta(u)$ close enough to $u$ – here, the restriction $d(x,u)<\delta/2$ suffices – and for such an $x$, a rational number $q>0$ such that $d(x,u)<q<\delta/2$, which works because the rationals are dense in the reals. It is not too hard to verify that this is enough for $u\in B_q(x)\subseteq B_{\delta}(u)\subseteq U$.
I concluded that $U\subseteq \bigcup \mathscr \{\ldots\}$ as above. Did I accidentally use the axiom of choice here?
All I found was Remark 1.3 in this paper, which states that the theorem holds in ZF. However it gives neither proof nor reference for that.
Let $x \in U$. Then for some $x \in X, \delta > 0$, $x \in B_\delta(x)$ (see comment above). Then consider $B_{\delta/2}(x) \neq \emptyset$. Taking our definition of dense to be "A set $D$ is dense if every nonempty open set contains a point of $D$", then $\exists p \in D$ s.t. $p \in B_{\delta/2}(x)$. Then there exists $q \in \mathbb{Q}$ s.t. $d(x,p) < q < \delta/2$ since $(d(x,p),\delta/2)$ is a nonempty open set in $\mathbb{R}$ and $\mathbb{Q}$ is dense in $\mathbb{R}$. Then you can easily verify $B_q(p) \subseteq B_\delta(x)$ and $x \in B_q(p)$. Then $B_q(p) \in \mathcal{S}$. Thus $\forall x \in U \exists V \in \mathcal{S} : s \in V \in \mathcal{S}$. Then $\cup \mathcal{S}$ is defined as $\{x \mid \exists V \in \mathcal{S}: x \in V\}$. From this immediately follows $\forall x \in U: x \in \cup \mathcal{S}$, or equivalently $U \subseteq \cup \mathcal{S}$.
I hope this more explicit proof shows AC is nowhere needed.