TITLE: Do we need full choice to "efficiently" use (sub)bases?
QUESTION [14 upvotes]: 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.)
EDIT: I do not require that $u\in f(u)$ (resp. $u\in\bigcap f(u)$) in the principles above, although that is a very natural requirement to include.

REPLY [10 votes]: EDIT: this answer addresses a variant of this question in which we require that $u\in f(u)$, respectively $u\in\bigcap f(u)$. Unfortunately it doesn't seem to say anything about the question as stated.
The subbase version implies AC. Let $X_i,i\in I$ be a family of nonempty sets. We may assume they are pairwise disjoint and are disjoint from $I$, and that $|X_i|>1$ for all $i$. Define a topology on $I\cup\bigcup_{i\in I}X_i$ by taking a subbasis $B$ consisting of sets $\{i,x\}$ for all $x\in X_i$. Then $I$ is an open subset in this topology: indeed, for any $i\in I$ and distinct $x,y\in X_i$ we have $\{i\}=\{i,x\}\cap\{i,y\}$.
Now any function $f:I\to[B]^{<\omega}$ such that $I=\bigcup_i(\bigcap f(i))$ and $i\in f(i)$ must satisfy that $f(i)$ consists of (finitely many) sets of the form $\{i,x\}$, and so $\bigcup f(i)\setminus\{i\}$ is a nonempty finite subset of $X_i$. We conclude that axiom of multiple choice holds, and hence so does axiom of choice.
Update: the basis version also implies (multiple) choice. Take the same set $X=I\cup\bigcup_{i\in I}X_i$ as above, but this time take the following basis of a topology: we take $B$ to consist of sets of the form $\{i\}\cup A$, where $A$ is a proper subset of $X_i$ whose complement in $X_i$ is finite. Any two elements of $B$ are either disjoint or their intersection is in $B$ as well. As $\bigcup B=X$, this implies $B$ is a basis of a topology. In particular $X$ itself is a union of elements of $B$. Let $f:X\to B$ be a function such that $x\in f(x)$ for all $x$. Then $X_i\setminus f(i)$ is a nonempty finite subset of $X_i$, giving multiple choice.