A Hausdorff topological space is said to be normal if given any two disjoint closed sets $A$ and $B$, there exist disjoint open sets $U$ and $V$ such that $A\subset U$ and $B \subset V$. Show that if $X$ is a normal space with at least two distinct points, then there exist non-constant real-valued continuous functions on $X$.
I have done an introductory course in topology. However I don't know where to start with this one. Any leads?
Here is a very explanatory proof of Urysohn's lemma which constructs a non-constant function for each pair of disjoint closed subsets of $X$.
Let $X$ be a Hausdorff normal space and $A$ and $B$ be two disjoint closed subsets of $X$. First prove that if for any closed set $A$ and an open set $U$ containing $A$, there is an open set $V$ such that $A \subset V \subset clV \subset U$. Consider the countable set $Q = [0,1]\cap \mathbb{Q}$. We will construct a sequence of open sets of $X$ to construct such a function. Since $Q$ is countable, we can construct a sequence $(p_n)_{n \in \mathbb{N}_{0}}$ that consists of all rationals in $Q$. W.l.o.g assume $p_0 = 0$ and $p_1 = 1$.
The sequence must satisfy for all $p < p'$ in $Q$, $clU_p \subset U_{p'}$. Define $U_1 = X\setminus B$. Note that $A \subset U_1$, so that we can find an open set $U_0$ containing $A$ and $U_1$ contains its closure. As $Q$ is countable, we can inductively define all such sets $U_p$ for $p \in Q$ as follows:
Suppose $U_0,U_1, U_{p_2}, U_{p_3}, \ldots U_{p_n}$ are given. Let $p_{n+1}$ be the next rational in the sequence. Then, since $\{p_i\ |\ 0\leq i \leq n+1\}$ is a well ordered set, $p_{n+1}$ has an immediate successor-and-predecessor. Say $p_i < p_{n+1} < p_j$. Then, we know that $clU_{p_i}\subset U_{p_j}$ and so by the same argument as above, we can find an open set $U_{p_{n+1}}$ such that $clU_{p_i}\subset U_{p_{n+1}}$ and $cl U_{p_{n+1}} \subset U_{p_j}$. By induction we can find all of these $U_{p_n}$ that satisfy the requirement.
Now, we extend $U_p$ for all rationals in $\mathbb{R}$ by letting $U_p = \emptyset$ if $p <0$ and $U_p = X$ if $p > 1$. You can check that $p \leq q$ in $\mathbb{Q}$ holds iff $cl U_p \subset U_q$.
For each $x \in X$, let $\mathbb{Q}(x): = \{p \in \mathbb{Q}\ |\ x \in U_p\}$. Note $\mathbb{Q}(x)$ is bounded below by $0$, so it makes sense to define $f:X\rightarrow [0,1]$ by $f(x) = inf \mathbb{Q}(x)$.
If $x \in A$, then $x \in U_p$ for each $p \in \mathbb{Q}^+$ so that $f(x)=0$ and if $x \in B$, then $\mathbb{Q}(x)$ consists of all rationals greater than $1$. Thus $f(x)=1$. Now we need only show that $f$ is continuous, so to that end we prove: Given $x$ in $X$ and $f(x) \in (c,d)$, we find an open set $U$ such that $f[U]\subset (c,d)$. Choose rational numbers $r,s$ in $Q$ such that $c<r<f(x) <s <d$. Then $U:= U_s\setminus clU_r$ is the desired open set.