Kunen exercise IV.4.13 (4): Topological version of effective AC I am dealing with Kunen's The Foundations of mathematics exercise IV.4.13 (4):
Let $X$ denote the Cantor set. Prove if $S\subset X\times X$ is open, then there is an $F\subset S$ such that $F$ is the graph of a function and $\text{dom} (F)=\text{dom} (S)$ and $F$ is continuous on $\text{dom} (F)$.
Kunen called this a topological version of the 'effective AC', with open sets analogous to $\Sigma_1$ sets, closed analogous to $\Pi_1$, and clopen analogous to $\Delta_1$, continuous analogous to computable in a zero-dimensional space. Thus, one natural approach is to imitate the proof of its recursion theory counterpart:
If $S\subset HF^2$ is $\Sigma_1$, then there is a partial computable function $F$ of 1 variables such that $F\subset S$ and $\text{dom} (F)=\text{dom} (S)$.
However, the proof makes use of a computable well-ordering on $HF$, for which I am unable to make connection to the topology version. The exercises proceeding it establish an analog of $\Sigma_1$ reduction (i.e. If $U,V\subset X$ open, then there are $U'\subset U$ and $V'\subset V$ such that $U'\cup V'=U\cup V$ and $U'\cap V' = \emptyset$.), though I am not sure if this will be of help.
 A: We'll take $X$ to be Cantor space: the topological space $2^\omega$ with basic open sets $\mathscr{N}_s = \{f\in X \mid f\upharpoonright\text{dom}(s)=s\}$ for $s\in 2^{<\omega}.$  (This is homeomorphic to the Cantor set with the subspace topology inherited from the reals.)
Let $S$ be an open subset of $X\times X.$
For any $f\in\textrm{dom}(S),$ define $i(f)$ to be the least $i\lt\omega$ such that for some $t\in 2^{\lt\omega},\; \mathscr N_{f\upharpoonright i}\times\mathscr N_t\subset S.$  (There exists an $i$ with this property since some $\langle f, g\rangle\in S$ and $S$ is open, and it follows that $\langle f, g\rangle$ is in some basic open set that is entirely contained in $S.)$
Using the usual enumeration of $2^{\lt\omega},$ let $t_f$ be the least $t\in 2^{\lt\omega}$ such that $\mathscr N_{f\upharpoonright i(f)}\times\mathscr N_t\subset S.$
Finally, define $F\colon\textrm{dom}(S)\to X$ by setting $F(f) = t_f\,^\smallfrown\,\langle 0, 0, \dots\rangle.$
Note that if $g$ is any member of $\textrm{dom}(S)$ with $g\upharpoonright i(f) = f\upharpoonright i(f),$ then $i(g) = i(f),$ so $t_f=t_g,$ and it follows that $F(f) = F(g).$
To see that $F$ is continuous, let $F(f)$ be in some open set $U.$ Then the range of $F$ on the open set $\mathscr{N}_{f\upharpoonright i(f)}\cap\textrm{dom}(S)$ (which $f$ belongs to) is equal to the singleton $\{F(f)\},$ which is entirely contained in $U.$

By the way, the enumeration of $2^{\lt\omega},$ which is tantamount to a counting of the basic open sets, is what corresponds to the well-ordering of $\mathsf{HF}$ in the recursion-theory proof OP was trying to emulate.
