Kunen exercise IV.4.13 (4): Topological version of effective AC

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.