Setup. Let $D = \mathbb{Q} \cap [0, 1]$ denote the rationals in $[0, 1]$. We will construct a family $\{U_r\}_{r \in D}$ of open sets satisfying:

• (i) $A \subseteq U_r$ for all $r \in D$,
• (ii) $U_r \subseteq X \setminus B$ for all $r \in D$,
• (iii) If $r < s$ (both in $D$), then $\overline{U_r} \subseteq U_s$.

We then define $f$ using these sets.

Step 1: Construction of $\{U_r\}$ for dyadic rationals.

Let $D_0 = \{k/2^n : 0 \leq k \leq 2^n, n \geq 0\} \cap [0, 1]$ be the dyadic rationals. We enumerate $D_0$ as $r_1, r_2, r_3, \ldots$ with $r_1 = 1$ and $r_2 = 0$.

Base: Set $U_1 = X \setminus B$. This is open and $A \subseteq U_1$ (since $A \cap B = \emptyset$).

Since $A$ is closed and contained in the open set $U_1$, and $X$ is normal, by the normality characterization there exists an open $U_0$ with: $A \subseteq U_0 \subseteq \overline{U_0} \subseteq U_1.$

Induction: Suppose we have defined $U_r$ for a finite set of dyadic rationals $r_1, \ldots, r_n$ satisfying condition (iii). Let $r_{n+1}$ be the next dyadic rational. Among $r_1, \ldots, r_n$, let $p$ be the largest value less than $r_{n+1}$ and $q$ the smallest value greater than $r_{n+1}$. Then $\overline{U_p} \subseteq U_q$ by the inductive hypothesis.

Apply normality: $\overline{U_p}$ is closed and contained in the open set $U_q$, so there exists an open $U_{r_{n+1}}$ with: $\overline{U_p} \subseteq U_{r_{n+1}} \subseteq \overline{U_{r_{n+1}}} \subseteq U_q.$

By induction, $\{U_r\}_{r \in D_0}$ satisfies (i), (ii), (iii) for all dyadic rationals.

Extension to all rationals. For any $r \in D$, we can extend by setting $U_r = \bigcup \{U_s : s \in D_0, s \leq r\}$. The nestedness property (iii) ensures this is consistent. (In practice, the dyadic rationals suffice for defining $f$.)

Step 2: Definition of $f$.

For each $x \in X$, define: $f(x) = \inf\{r \in D_0 : x \in U_r\},$ with the convention that $\inf \emptyset = 1$.

Properties:

• If $x \in A$: then $x \in U_r$ for all $r \in D_0$ (by (i)), so $f(x) = \inf D_0 = 0$.
• If $x \in B$: then $x \notin U_r$ for all $r \in D_0$ (by (ii)), so $f(x) = 1$.
• For all $x$: $0 \leq f(x) \leq 1$.
• $f(x) < r \iff x \in U_r$ for some $r' < r$ in $D_0$ $\iff$ $x \in \bigcup_{s < r, s \in D_0} U_s$.
• $f(x) \leq r \iff x \in U_s$ for all $s > r$ in $D_0$ $\iff$ $x \in \bigcap_{s > r, s \in D_0} U_s$. By (iii), this implies $x \in \overline{U_r}$.

Step 3: Continuity of $f$.

It suffices to show that $f^{-1}([0, a))$ and $f^{-1}((a, 1])$ are open for every $a \in (0, 1)$, as these generate the topology on $[0, 1]$.

Claim: $f^{-1}([0, a)) = \bigcup_{r < a, r \in D_0} U_r$.

If $f(x) < a$, then $\inf\{r : x \in U_r\} < a$, so $x \in U_r$ for some $r < a$. Conversely, if $x \in U_r$ for some $r < a$, then $f(x) \leq r < a$. This set is open (union of open sets).

Claim: $f^{-1}((a, 1]) = \bigcup_{r > a, r \in D_0} (X \setminus \overline{U_r})$.

If $f(x) > a$, then $x \notin U_r$ for all $r \leq a$. Choose $r \in D_0$ with $a < r < f(x)$. Then $x \notin U_r$, so $x \notin \overline{U_s}$ for some $s$ with $a < s < r$ (since $\overline{U_s} \subseteq U_r$). Thus $x \in X \setminus \overline{U_s}$.

Conversely, if $x \in X \setminus \overline{U_r}$ for some $r > a$, then $x \notin U_s$ for any $s \leq r$ (since $U_s \subseteq \overline{U_s} \subseteq U_r$ for $s < r$ and $U_r \subseteq \overline{U_r}$). Wait -- more carefully: $x \notin \overline{U_r}$ implies $x \notin U_r$. For $s < r$, $\overline{U_s} \subseteq U_r \subseteq \overline{U_r}$, so $x \notin U_s$ either. Thus $f(x) \geq r > a$.

This set is a union of open sets (each $X \setminus \overline{U_r}$ is open), hence open.

Therefore $f$ is continuous.