First we construct a family $U_p$ of open sets of $X$ indexed by the rationals such that if $ p < q$ , then $\bar{U_p} \subseteq U_q$ . These are the sets we will use to define our continuous function.

Let $P = \mathbb{Q} \cap [0,1]$ . Since $P$ is countable, we can use induction (or recursive definition if you prefer) to define the sets $U_p$ . List the elements of $P$ is an infinite sequence in some way; let us assume that $1$ and $0$ are the first two elements of this sequence. Now, define $U_1 = X \backslash D$ (the complement of $D$ in $X$ ). Since $C$ is a closed set of $X$ contained in $U_1$ , by normality of $X$ we can choose an open set $U_0$ such that $C \subseteq U_0$ and $\bar{U_0} \subseteq U_1$ .

In general, let $P_n$ denote the set consisting of the first $n$ rationals in our sequence. Suppose that $U_p$ is defined for all $p \in P_n$ and \begin{equation} \textrm{if} \ p < q, \textrm{then} \ \bar{U_p} \subseteq U_q. \end{equation}Let $r$ be the next rational number in the sequence. Consider $P_{n+1} = P_n \cup \{r\}$ . It is a finite subset of $[0,1]$ so it inherits the usual ordering $<$ of $\mathbb{R}$ . In such a set, every element (other than the smallest or largest) has an immediate predecessor and successor. We know that $0$ is the smallest element and $1$ the largest of $P_{n+1}$ so $r$ cannot be either of these. Thus $r$ has an immediate predecessor $p$ and an immediate successor $q$ in $P_{n+1}$ . The sets $U_p$ and $U_q$ are already defined by the inductive hypothesis so using the normality of $X$ , there exists an open set $U_r$ of $X$ such that

We have defined $U_p$ for all rationals in $[0,1]$ . Extend this definition to every rational $p \in \mathbb{R}$ by defining

Now, given $x \in X$ , define $\mathbb{Q}(x) = \{p : x \in U_p\}$ . This set contains no number less than $0$ and contains every number greater than $1$ by the definition of $U_p$ for $p < 0$ and $p > 1$ . Thus $\mathbb{Q}(x)$ is bounded below and its infimum is an element in $[0,1]$ . Define

Finally we show that this function $f$ we have defined satisfies the conditions of lemma. If $x \in C$ , then $x \in U_p$ for all $p \ge 0$ so $\mathbb{Q}(x)$ equals the set of all nonnegative rationals and $f(x) = 0$ . If $x \in D$ , then $x \notin U_p$ for $p \le 1$ so $\mathbb{Q}(x)$ equals all the rationals greater than 1 and $f(x) = 1$ .

To show that $f$ is continuous, we first prove two smaller results:

(a) $x \in \bar{U_r} \Rightarrow f(x) \le r$

Proof. If $x \in \bar{U_r}$ , then $x \in U_s$ for all $s > r$ so $\mathbb{Q}(x)$ contains all rationals greater than $r$ . Thus $f(x) \le r$ by definition of $f$ .

(b) $x \notin U_r \Rightarrow f(x) \ge r$ .

Proof. If $x \notin U_r$ , then $x \notin U_s$ for all $s < r$ so $\mathbb{Q}(x)$ contains no rational less than $r$ . Thus $f(x) \ge r$ .

Let $x_0 \in X$ and let $(c,d)$ be an open interval of $\mathbb{R}$ containing $f(x)$ . We will find a neighborhood $U$ of $x_0$ such that $f(U) \subseteq (c,d)$ . Choose $p,q \in \mathbb{Q}$ such that

Finally, let $x \in U$ . Then $x \in U_q \subseteq \bar{U_q}$ , so $f(x) \le q$ by (a). Also, $x \notin \bar{U_p}$ so $x \notin U_p$ and $f(x) \ge p$ by (b). Thus