Assuming Tietze's Extension Theorem prove Urysohn's lemma This question is from Wayne Patty's Topology section 5.5 and I was unable to solve it.
So, I am looking for help here.

Assuming Tietze Extension Theorem prove Urysohn's Lemma.

Urhysohn's Lemma states that A $T_1$ space is normal iff for each pair A,B of disjoint closed subsets of X there is a continuous function $f: X \to I$ such that f(x)=0 for all $x\in A$ and f(x)=1 for all $x\in B$.
Now assuming that $T_1$ space is normal. Let C, D be disjoint closed subsets then there exists disjoint U and V such that $C\subseteq U$ and $D\subseteq V$ . I have to find a function f satisfying required conditions . But Tietze theorem doesn't guarantess existence of such function and I am not able to construct it.
For the other side, if C and D are disjoint closed sets I am given continuous function f such that f(x) =0 for all $x\in C$ and f(x)=1 for all $x\in D$ but how to prove existence of such open sets?
I am completely stumped and need help.
 A: This very easy. If $C$ and $D$ are disjoint closed sets then the function $g: C \cup D \to I$ defined by $g(x)=0$ for $x \in C$ and $g(x)=1$ for $x \in D$ is a continuous function. By Tietze Theorem this extends to a  continuous function $f$ on $X$ and this finishes the proof.
[If your version of Tietze's Theorem has codomain as $\mathbb R$ then you can look at $\max \{0, \min \{f(x),1\}\}$ at the end of my answer].
[To prove that $g$ is continuous show that inverse image of any closed set in $I$ is closed in $C \cup D$ (in the subspace topology)].
A: You misinterpret the question:
"Urysohn's lemma for $X$" means

For all $A,B \subseteq X$ where $A$ and $B$ are closed and disjoint there exists a continuous $f: X \to [0,1]$ with $f[A] \subseteq \{0\}$ and $f[B] \subseteq \{1\}$.

"Tietze's extension theorem for $X$" means

For all $A \subseteq X$ closed and all continuous $f: A \to \Bbb R$ there exists a continuous $F: X \to \Bbb R$ so that $F\restriction_A = f$.

The question that your text asks is really:

For all spaces $X$, show that "Tietze's extension theorem for $X$" implies "Urysohn's lemma for $X$".

So whether $X$ is normal or $T_1$ or nor matters not a whit.
