Equivalence of countable choice for subsets of the reals and "second countable $\implies$ Lindelöf" Looking for a proof that in a second countable space every open cover has a countable subcover -- i.e. every s.c. space is a Lindelöf space -- I bumped into this question. That answered my question. I then left a couple of comments on both the then present answers (which are, to this date, the only ones) and received this reply, stating the axiom of countable choice for subsets of the reals is actually equivalent to every s.c. space being Lindelöf. Now it seems incredible that a special case of the axiom of countable choice, which in its general form states any countable collection of sets has a choice function, so for all countable collection of sets $A_n$ there is a function associating to each set an element of that set, provided of course the sets are non-empty, and in its particular form says so for sets which are subsets of the reals, be equivalent to s.c. implies Lindelöf. At most, I would expect the general form to be equivalent to that, since that s.c. implies Lindelöf is proven with a form of countable choice. So how can it be proved that the two results are equivalent?
This is not that incredible if you understand what's going on.

Given a second-countable space $X$, fix a countable basis $\{U_n\mid n\in\Bbb N\}$. Now you can observe that every open set is uniquely determined by the basis elements it contains, namely the function $U\mapsto\{n\in\Bbb N\mid U_n\subseteq U\}$ is an injective function. And since the real numbers can be thought of as subsets of $\Bbb N$, this is an injective function from the topology of $X$ into $\Bbb R$.

So if we are given an open covering, $\{V_i\mid i\in I\}$ we may assume that $|I|\leq|\Bbb R|$ by the above consideration, and for each $n\in\Bbb N$ we can look at the set $A_n=\{V_i\mid U_n\subseteq V_i\}$. 

Not every $A_n$ is non-empty, but at least one of them is, of course. By the translation between open sets and real numbers, this is reduced to a countable family of sets of reals. Using $\sf AC_\omega(\Bbb R)$ we can choose from each non-empty $A_n$, and we can prove that this is a countable subcover by noting that $\{U_n\mid A_n\neq\varnothing\}$ is an open cover as well.



On the other hand, if $\{A_n\mid n\in\Bbb N\}$ is a family of pairwise disjoint sets of reals without a choice function. We may assume that each $A_n$ is a subset of $[n,n+1)$ by using canonical bijections of $[\inf A_n,\sup A_n+1)$ and $[n,n+1)$.

Now consider the open covering of $\Bbb R$ defined by $\{(-r,r)\mid\exists n: r\in A_n\}$. Since the $A_n$'s grow arbitrarily large this is an open covering indeed, so by the assumption it has a countable subcover, which gives for each $A_n$ a countable subset, so by diluting this subcover we may assume that each $A_n$ gives at most a single real number.

And therefore we found a partial choice function for the $A_n$'s. Now it is enough to show that it is indeed enough to consider partial choice functions.

Indeed recall that $\Bbb R^{&lt;\omega}$, the finite sequences of real numbers have the same cardinality as $\Bbb R$ itself. Given a sequence of sets $X_n$ which are non-empty, consider $A_n=\prod_{k\leq n}X_n$, then the $A_n$'s are non-empty sets of finite sequences of real numbers, so by uniform encoding we may assume these are sets of real numbers. If $I\subseteq\Bbb N$ is infinite and $\sigma$ is a function choosing from $\{A_i\mid i\in I\}$, then it defines a choice function from the $X_n$ by looking at $i=\min\{j\in I\mid j&gt;n\}$ and looking at $\sigma(i)(n)$ (recall that $\sigma(i)$ is a choice function which is defined on $X_n$).
Let $\mathcal B$ be the countable base of $X$. Then for any open cover $\mathcal U=\{U_\alpha:\alpha \in \Omega\}$, then for any $x\in X$ and $x \in U_\alpha$, there exists $B_x\in \mathcal B$ such that $x\in B_x \subset U_\alpha$. Then $\{B_x:x\in X\}$ is the subcover and countable.