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Urysohn Lemma:  If $X$ is normal then for any $A, B$ disjoint closed sets in $X$, there exists a
continuous function $f: X \rightarrow [0, 1]$ such that $f(A) = \{0\}$ and $f(B)  = \{1\}$

Proof:  Suppose $A, B$ disjoint closed sets in $X$,

Choose a bijective function $g: {\cal N} \rightarrow {\cal Q} \cap [0, 1]$ such that $g(1) = 1$ and 
$g(2) = 0$.
Let $g(n) = q_n$

Define $U_i = X$ for all $i > 1$
Define $U_1 = X - B$

Since $X$ is normal, there exists an open set $U_0$ such that $A \subset U_0 \subset 
\overline{U_0} \subset U_1$.

Similarly there exists an open set $U_{q_3}$ such that

\centerline{ $\overline{U_0} \subset U_{q_3} \subset 
\overline{U_{q_3}} \subset U_1$.}

Suppose $U_{q_1}, ..., U_{q_{n-1}}$ have been defined such that $U_{q_i}$ is open
and if $q_i < q_j$ then $\overline{U_{q_i}} \subset U_{q_j}$


Define $U_{q_n}$: 
\hfil \break  
Let $s = min\{ q_i < q_n ~|~ i = 1, ..., n-1\}$
\hfil \break  
Let $t = max\{ q_i > q_n ~|~ i = 1, ..., n-1\}$

Since $X$ is normal, there exists an open set $U_{q_n}$ such that $\overline{U_{s}} \subset U_{q_n} 
\subset
\overline{U_{q_n}} \subset U_t$.

Thus if $q_n < q_i$, then $q_n < t \leq q_i$.  
\v
Thus $\overline{U_{q_n}} \subset U_t \subset 
\overline{U_{t}} \subset U_{q_i}$

Thus if $q_i < q_n$, then $q_i \leq s < q_n$. 
\v
 Thus $\overline{U_{q_i}} \subset U_s \subset 
\overline{U_{s}} \subset U_{q_n}$

Hence we have defined $U_p$ for all $p \in {\cal Q} \cup [0, \infty)$ such that $U_p$ is open for all 
$p$ and if $p < 
q$ then $\overline{U_{p}} \subset U_{q}$.

Define $f: X \rightarrow {\cal R}$ by 
\hfil \break
$f(x) = inf\{p ~|~ x \in U_p,~ p \in {\cal Q} \cap [0, \infty) \}$

If $x \in A$, then $x \in U_0$.  
\hfil \break
Thus $f(x) = inf\{p ~|~ x \in U_p,~ p \in {\cal Q} \cap [0, \infty) \} = 0$

If $x \in B$, then $x \in U_p = X$ for all $p > 1$.
\hfil \break
But $x \not\in U_1 = X - B$. 
\hfil \break
Since $\overline{U_q}\subset U_1$ for all $q < 1$, $x \not\in U_q$ for all $q < 1$.  
\hfil \break
Thus $f(x) = 
inf\{p ~|~ x \in 
U_p,~ p \in {\cal Q} \cap [0, \infty) \} = 1$


Since $x  \in U_p = X$ for all $p > 1$. 
\hfil \break
$f(x) =
inf\{p ~|~ x \in
U_p,~ p \in {\cal Q} \cap [0, \infty) \} \leq 1 ~\forall x \in X$.

Since 
$inf\{p ~|~ ,~ p \in {\cal Q} \cap [0, \infty)
 \} = 0$,
\hfil \break
$inf\{p ~|~ x \in
U_p,~ p \in {\cal Q} \cap [0, \infty) \} \geq 0$ for all $x \in X$.

Thus $f(X) \subset [0, 1]$ and hence $f: X \rightarrow [0, 1]$ is continuous if and only if $f: 
X 
\rightarrow {\cal R}$ is continuous.

Claim:  $f: X \rightarrow {\cal R}$ is continuous.

Take $x \in X$ and $(a, b) \subset  {\cal R}$ such that $f(x) \in (a, b)$.

Take $p, q \in {\cal Q}$ such that $a < p < f(x) < q < b$

Claim:  $f(U_q - \overline{U_p}) \subset (a, b)$ and $x \in U_q - \overline{U_p}$

\end