Question:
Given two distinct points $x_1$, $x_2$ of normal space $X$, show that there exists continuous function $f:X\to\mathbb R$ such that $f(x_1)\neq f(x_2)$.
My attempt:
I thought last process of proof is using Urysohn's lemma, so I tried hard to take two closed sets $C$, $D$, which contains $x_1, x_2$. So long time I can't solved. Please help!
Your idea is good. Take $C=\{x_1\}$ and $D=\{x_2\}$, which are closed sets since, by definition, every normal space is $T_1$.