This question was previously asked and bountied at MSE without success.

Suppose $(X,\tau)$ is a topological space, $B$ is a base for $\tau$, and $U\in \tau$ is an open set. Consider the following two strategies for writing $U$ as a union of elements of $B$:

• We have $U=\bigcup\{V\in B: V\subseteq U\}$.

• For each $u\in U$ pick some $V_u\in B$ with $u\in V_u\subseteq U$; then $U=\bigcup\{V_u: u\in U\}$.

The first strategy has the advantage of not requiring the axiom of choice. If we pay attention to the number of basic opens required, however, it is noticeably inefficient: the first strategy might involve as many as $2^{\vert U\vert}$-many basic open sets, while the second involves at most $\vert U\vert$-many.

It's not hard to show that in fact this drop in efficiency is unavoidable: it is consistent with $\mathsf{ZF}$ that there is a space $(X,\tau)$, a base $B$ for $\tau$, and an open set $U\in\tau$ such that there is no map $f:U\rightarrow B$ with $\bigcup_{u\in U}f(u)=U$. I'm interested in the exact strength of the corresponding efficiency principle, as well as its "subbase" variation:

Over $\mathsf{ZF}$, are either of the following statements equivalent to $\mathsf{AC}$?

• For every topological space $(X,\tau)$, every base $B$ for $\tau$, and every $U\in\tau$, there is some $f:U\rightarrow B$ with $\bigcup_{u\in U}f(u)=U$.
• For every topological space $(X,\tau)$, every subbase $B$ for $\tau$, and every $U\in\tau$, there is some $f:U\rightarrow [B]^{<\omega}$ with $\bigcup_{u\in U}(\bigcap f(u))=U$.

(Above, "$[A]^{<\omega}$" denotes the set of finite subsets of $A$. So the subbase version of the principle is saying that we can write $U$ as the union of $U$-many finite intersections of subbase elements.)