show that the Tietze extension theorem implies the urysohn lemma The theorem states

If a continuous map $f\,:\,A\to\Bbb R$ with $A$ a closed subset of the
  normal topological space $X$ carries the standard topology, some
  continuous extension $F\,:\,X\to\Bbb R$ of $f$ satisfies $\sup\{|f(a)|\mid a\in A\}=\sup\{|F(X)|\mid x\in X\}$.

The lemma states

If for subsets $A,\,B$ of a topological space $X$ some continuous Urysohn function $F\,:\,X\to[0,\,1]$ has $\forall (a,\,b)\in A\times B((f(a),\,f(b))=(0,\,1))$, the same is true of the closures of $A,\,B$; in particular there exist respective neighbourhoods $U,\,V$ of $A,\,B$ with $U\cap V=\emptyset$.

I came across this problem in Munkres Topology second edition. I am having a hard time with this problem and would like some help on it. Would just writing out the proof for Urysohn's Lemma be enough to solve this problem? Or does it require something else?
 A: It's clear that the Tietze extension theorem implies Urysohn's lemma: if $A$ and $B$ are disjoint closed sets of a normal space $X$, define $f: A \cup B \to \mathbb{R}$ by $$f(x)=\begin{cases}0 & x \in A\\
1 & x \in B\end{cases}$$
and note that $A\cup B$ is closed in $X$ and $f$ is continuous (the gluing lemma for closed sets implies this) and so it has a continuous extension $F: X \to \mathbb{R}$ (forgetting the $\sup$-property, that we don't need) and this extension is a Urysohn function for $A$ and $B$ by definition. 
A: Suppose $a\in \overline A$ and $F(a)=r >0$. Then, there is a neighborhood $N$ of $a$ such that $x\in N\Rightarrow F(x)>r/2.$ But $N$ contains a point $y\in A$ and as $F(y)=0,$ we get a contradiction. We conclude that $F(a)=0.$
Similarly, $b\in \overline B\Rightarrow F(b)=1.$
This shows also that $\overline A\cap \overline B=\emptyset.$ 
So, from now on, we may assume $A$ and $B$ are closed and disjoint. 
Then, we have
$F(x)=\begin{cases} 0 &x\in A\\
1 & x\in B \end{cases}$.
$F$ is continuous on $A\cup B$, which is closed in $X$, so Tietze says there is a continuous extension of $F$ to all of $X$ such that $\sup\{f(x):x\in X\}\le 1$.
To finish, note that $U=\{F>1/2\}$ and $V=\{F<1/2\}$ are open, disjoint and contain $A$ and $B$, respectively.   
