Urysohn's Lemma: Proof Given a normal space $\Omega$.

Then closed sets can be separated continuously:
  $$h\in\mathcal{C}(\Omega,\mathbb{R}):\quad h(A)\equiv0,\,h(B)\equiv1\quad(A,B\in\mathcal{T}^\complement)$$
  Especially, it can be chosen as a bump: $0\leq h\leq1$

Though the idea is very clear it can be strikingly technical.
The goal here is to crystalize out the most lucid version!
Disclaimer: This is a record for future threads.
 A: $\newcommand{\cl}{\operatorname{cl}}$I prefer a more topological proof.
Let $D$ be the set of dyadic rationals in $[0,1]$. We shall construct a family $\{U(d):d\in D\}$ of sets $A \subseteq U(d) \subseteq \Omega\setminus B$, all open with the possible exception of $U(0)$, satisfying the condition that $\cl U(d_0)\subseteq U(d_1)$ whenever $d_0,d_1\in D$ and $d_0<d_1$. 
For $n\in\Bbb Z^+_0$ = set of non-negative integers let $D_n  = \{\frac{k}{2^n} : k\in\{0,\ldots,2^n\} \}$. We have $D_{n} \subseteq D_{n+1}$ and $\bigcup_{n=0}^\infty D_n = D$. We shall construct the desired $U(d)$ inductively for $d \in D_n$.
For $n = 0$ let $U(0)=A$ and $U(1)=\Omega\setminus B$. Note that $U(0)$ is closed, but in general not open.
Now suppose the $U(d)$ have been constructed for $d \in D_n$. By normality, for each $k\in\{0,\ldots,2^n\}$ there is an open $U\left(\frac{2k+1}{2^{n+1}}\right)$ such that
$$\cl U\left(\frac{k}{2^n}\right)=\cl U\left(\frac{2k}{2^{n+1}}\right)\subseteq U\left(\frac{2k+1}{2^{n+1}}\right)\subseteq\cl U\left(\frac{2k+1}{2^{n+1}}\right)\subseteq U\left(\frac{k+1}{2^n}\right)\;.$$
Let
$$h:\Omega\to[0,1]:x\mapsto\begin{cases}
1,&\text{if }x\notin U(1)\\
\inf\{d\in D:x\in U(d)\},&\text{otherwise}\;.
\end{cases}$$
Clearly $h[A]=\{0\}$ and $h[B]=\{1\}$, so it only remains to show that $h$ is continuous. First note that if $r\in D$ and $x\in U(r)$, then $h(x)\le r$, while if $x\in\Omega\setminus\cl U(r)$ then $h(x)\ge r$. Now let $x\in\Omega$ and $\epsilon>0$ be arbitrary. 


*

*If $h(x)=0$, choose $r\in D$ so that $0<r<\epsilon$; then $V=U(r)$ is an open nbhd of $x$ such that $h[V]\subseteq[0,\epsilon)$.

*If $h(x)=1$, choose $r\in D$ so that $1-\epsilon<r<1$; then $V=\Omega\setminus\cl U(r)$ is an open nbhd of $x$ such that $h[V]\subseteq(1-\epsilon,1]$.

*If $0<h(x)<1$, choose $r,s\in D$ so that $h(x)-\epsilon<r<h(x)<s<h(x)+\epsilon$; then $V=U(s)\setminus\cl U(r)$ is an open nbhd of $x$ such that $h[V]\subseteq(h(x)-\epsilon,h(x)+\epsilon)$.
Thus, $h$ is continuous at every point of $\Omega$.
A: Disclaimer: This is community wiki; feel free to correct, edit, expand etc.! ;)
(This is a slight modification taken from Rudin, Real and Complex Analysis.)
Regard for simplicity:
$$C:=B\subseteq A^\complement=:U$$
Denote for construction the rationals:
$$S:=\mathbb{Q}\cap[0,1]$$
By normality one can find inductively open $V_s$ for $s \in S$ such that
$$C\subseteq V_1, \quad \overline{V_0}\subseteq U,\quad V_s\subseteq\overline{V_{s'}} \text{ for } s>s' .$$
Consider the upper and lower semicontinuous functions:
$$\underline{h}_s:=s\chi_{V_s}:\quad\underline{h}_s^{-1}(a,\infty)=\varnothing,V_s,\Omega\in\mathcal{T}$$
$$\overline{h}_s:=(1-s)\chi_{\overline{V_s}}+s:\quad\overline{h}_s^{-1}(-\infty,b)=\varnothing,\overline{V_s}^\complement,\Omega\in\mathcal{T}$$
Pointwise suprema resp. infima preserve lower resp. upper semicontinuity:
$$\underline{h}:=\sup_{s\in S}\underline{h}_s:\quad\underline{h}^{-1}(a,\infty)=\{\omega:\underline{h}(\omega)>a\}=\bigcup_{s\in S}\{\omega:\underline{h}_s(\omega)>a\}=\bigcup_{s\in S}\underline{h}_s^{-1}(a,\infty)\in\mathcal{T}$$
$$\overline{h}:=\inf_{s\in S}\overline{h}_s:\quad\overline{h}^{-1}(-\infty,b)=\{\omega:\overline{h}(\omega)<b\}=\bigcup_{s\in S}\{\omega:\overline{h}_s(\omega)<b\}=\bigcup_{s\in S}\overline{h}_s^{-1}(-\infty,b)\in\mathcal{T}$$
They approach each other as by contradiction:
$$\underline{h}(\omega)<s<s'<\overline{h}(\omega)\implies\omega\in\overline{V_{s'}}\subseteq V_s\not\owns\omega$$
$$\underline{h}_s(\omega)>\overline{h}_{s'}(\omega)\implies\omega\in V_s\subseteq\overline{V_s}\subseteq V_{s'}\subseteq\overline{V_{s'}}\not\owns\omega\quad(s>s')$$
So they together become continuous:
$$h:=\overline{h}=\underline{h}:\quad h^{-1}(c-\varepsilon,c+\varepsilon)=h^{-1}(-\infty,c+\varepsilon)\cup h^{-1}(c-\varepsilon,\infty)\in\mathcal{T}$$
Thus the limiting function is the desired bump:
$$h(C)\subseteq h(V_1)\equiv1,\,h(U^\complement)\subseteq h(\overline{V_0}^\complement)\equiv0$$
A: (Various versions of Urysohn’s lemma)
Dugundji = Dugundji, J.: Topology, Boston: Allyn and Bacon, 1966.
Pervin = Pervin, W.J.: Foundations of General Topology, New York, Academic Press, 1964.
Pontryagin = Pontryagin, L.: Topological Groups, translated from the Russian by Emma Lehmer, Princeton: Princeton
University Press, 1946.
Rudin = Rudin, W.: Real and Complex Analysis, 2nd ed., New York: McGraw-Hill, 1974.
A. Pontryagin [p.46, l.$-$16]: $R$ is a compact regular topological space satisfying the second axiom of countability; the key ideas: Pontryagin [p.46, l.$-$11--l.$-$8].
B. Dugundji [p.146, Theorem 4.1 (1) $\Rightarrow$ (2)]: $Y$ is normal.
C. Pervin [p.158, Urysohn's lemma]: $X$ is normal.
D. Rudin [p.40, l.14]: $X$ is a locally compact Hausdorff space.
Remark 1. The proof of A is the primitive model. Regular Lindel$\ddot{\text{o}}$f $\Rightarrow$ Normal [Dugundji [p.311, diagram]], so $B\Rightarrow A$. Thus, B is stronger than A. A stronger hypothesis implies stronger result, so $\bar{G}_r\cap F=\varnothing$ [Pontryagin [p.47, l.8]] is stronger than $U(r)\cap B=\varnothing$  [Dugundji [p.147, l.13]]. The proof method given in A and B is the same.
Remark 2. We may generalize the construction given in Pervin [p.158, l.$-$12--l.$-$2] by replacing $\{r_n\in \mathbb{N}|n\ge 3\}$ with any countable dense set $D$ of real numbers in $(0,1)$. Then the construction given in Dugundji [p.147, l.10--l.26] will be a special case of this generalized construction.
Remark 3. Locally compact $\Rightarrow$ Completely regular [Dugundji [p.311, diagram]], so we cannot use B or C to prove D. A remedy is to require that one of the two closed sets be compact. Rudin [p.38, Theorem 2.7] corresponds Dugundji [p.144, 3.2(2)]. The construction given in Rudin [p.40, l.$-$9--p.41, l.4] is similar to that given in Pervin [p.158, l.$-$12--l.$-$2] except that $r<s \Rightarrow c(G_r)\subset G_s$ [The larger a rational number is, the larger the corresponding set is; Pervin [p.158, l.$-$11]] is different from Rudin [p.41, (3)][The larger a rational number is, the smaller the corresponding set is]. The continuity of $\mathfrak{g}$ [Pervin [p.158, l.$-$1; p.159, l.9]] is proved by a direct argument, while the continuity of $f$ [Rudin [p.41, l.5--l.12]] is proved by an indirect [through semiconcontinuity] and more generalized argument; the direction of generalization: continuity $\rightarrow$ semicontinuity.
For the applications of Urysohn’s lemma, see Example 6.169 and Example 6.170 in https://sites.google.com/view/lcwangpress/%E9%A6%96%E9%A0%81/papers/mathematical-methods.
