Since the post is quite long I immediately precise that I substantially ask to prove the inclusion $\eqref{eq:simple3} $ using the inclusion $ \eqref{eq:simple2} $ and if you want you may not read the marked parts at a first look: anyway it is possibile that $\eqref{eq:simple3}$ does not holds so that in this case I ask how to prove the continuity of $\eqref{eq:simple4}$. Finally, I point out I did not study uniform spaces so that I courteously ask to do not use uniformity theory but I ask simply to use Munkres hint as showed to follow: I apologise for my ignorance, sorry really. So let's we well illustrate the issue.
Definition
A topological group is a group $(X,*)$ equipped with a topology $\cal T$ with respect to which the functions $$ p:X\times X\ni (x_1,x_2)\longrightarrow x_1*x_2\in X\quad\text{and}\quad s:X\ni x\longrightarrow x^{-1}\in X $$ are continuous.
So with respect the last definition I am trying to prove that any topological group $(X,*,\cal T)$ is completely regular using the following Munkres procedure.
So first of all let's we assume that for any closed set $C$ disjoint from the identity $e$ there exists a continuous function $f$ from $X$ to $[0,1]$ such that $$ f(e)=0\quad\text{and}\quad f[C]\subseteq\{1\} $$ and thus let's we prove that the statement woulds holds.
So if $x_0\in X$ is different from $e$ then we observe that the function $$ f_0: X\ni x\longrightarrow x*x_0^{-1}\in X $$ is a homeomorphism such that $$ f_0(x_0)=e $$ so that if $x_0\notin C$ then by assumption there exists a continuous function $f$ from $x$ to $[0,1]$ such that $$ f(e)=0\quad\text{and}\quad f\big[f_0[C]\big]\subseteq\{1\} $$ and thus finally $f\circ f_0$ is a continuous function from $X$ to $[0,1]$ such that $$ (f\circ f_0)(x_0)=0\quad\text{and}\quad(f\circ f_0)[C]\subseteq\{1\} $$ which proves that the complete regularity of $X$ follows proving the existence of a function $f$ from $X$ to $[0,1]$ which separated $e$ from any closed set not containing it.
Now it is a well know result that for any neighborhood $U$ of $e$ there exists a neighborhood $V$ of $e$ such that $$ V^2\subseteq U $$ so that let's we make a sequence $(V_n)_{n\in\omega}$ of neighborhood of $e$ such that $$ V_0\subseteq X\setminus C $$ and such that $$ V_{n+1}*V_{n+1}\subseteq V_n $$ for each $n\in\omega$. Now let be $$ \Delta:=\Big\{\frac m{2^n}:m,n\in\Bbb Z\Big\} $$ the set of dyadic numbers which is countable and dense: so let's we put $$ \begin{equation}\tag{1}\label{eq:simple1}U_{\frac m {2^n}}:=\begin{cases}\emptyset,\,\text{if }\frac m {2^n}\le 0\\V_{n},\,\text{if } \frac m {2^n}=\frac 1{2^n}\le 1\\V_{n+1}*U_{\frac h{2^n}}\,\text{if } \frac m {2^n}=\frac{2h+1}{2^{n+1}}<1\\ X,\,\text{if }\frac m {2^n}>1\end{cases}\end{equation} $$ for each $m,n\in\Bbb Z$. Therefore let's we verify that $U_\delta$ is defined for each $\delta\in\Delta$ but clearly $U_\delta$ is surely defined and is open when $\delta\notin(0,1)$ so that we have only to prove that $U_\delta$ is defined for $\delta\in(0,1)$.
So we observe that $2^n$ is positive for each $n\in\Bbb Z$ so that the inequality $$ 0<\frac m{2^n}<1 $$ for $m,n\in\Bbb Z$ holds if and only if also the inequality $$ 0<m<2^n $$ holds but if $n\le 0$ then $2^n$ is less than $1$ so that the last inequality can holds only for $n>0$ and thus let's we prove inductively over $n$ that $U_{\frac m{2^n}}$ is well defined and is open for any $m<2^n$: so if $n=1 $ then $m$ can only be equal to $1$ and $U_{\frac 1 2}$ is equal to $V_1$ so that let's we assume that the statement holds for any $n$; thus we observe that if $m<2^{n+1}$ is even then there exists $h\in\Bbb N$ such that $$ U_{\frac m{2^{n+1}}}=U_{\frac{2h}{2^{n+1}}}=U_{\frac h {2^n}} $$ so that by the inductively step $U_{\frac m{2^{n+1}}}$ is defined and is open if $h<2^n$ - but also if $2^n\le h$; otherwise if $m<2^{n+1}$ is odd then there exists $h\in\Bbb N$ such that $$ U_{\frac m{2^{n+1}}}=U_{\frac {2h+1}{2^{n+1}}}=V_{n+1}*U_{\frac h{2^n}} $$ but by the inductively set $U_{\frac h{2^n}}$ is defined and is open (it is indeed product of open sets!) if $h<2^n$ - and also if $2^n\le h$ - so that we conclude that $U_{\frac m{2^{n+1}}}$ is defined.
Now if $n$ is negative then $\frac m{2^n}$ is greater than $1$ or less than $0$ so that the inclusion $$ \begin{equation}\tag{2}\label{eq:simple2}V_n*U_{\frac k{2^n}}\subseteq U_{\frac{k+1}{2^n}}\end{equation} $$ trivially holds ans thus let's we prove inductively that it holds also for $n$ positive assuming that the inequality $$ 0<k<2^n $$ holds because otherwise $\eqref{eq:simple2}$ trivially holds.
So first of all we observe that $\eqref{eq:simple2}$ holds trivially for $n=0$. Therefore let's we assume that the inclusion holds for any $n\in\Bbb N$. So we observe that if $k$ is even then there exists $h\in\Bbb N$ such that $$ V_{n+1}*U_{\frac k{2^{n+1}}}=V_{n+1}*U_{\frac{2h}{2^{n+1}}}=V_{n+1}*U_{\frac h{2^n}}=U_{\frac{2h+1}{2^{n+1}}}=U_{\frac{k+1}{2^{n+1}}} $$ where we used $\eqref{eq:simple1}$; otherwise if $k$ is odd then there exist $h\in\Bbb N$ such that $$ V_{n+1}*U_{\frac k{2^{n+1}}}=V_{n+1}*U_{\frac{2h+1}{2^{n+1}}}=\\V_{n+1}*V_{n+1}*U_{\frac h{2^n}}\subseteq V_n*U_{\frac h{2^n}}\subseteq U_{\frac{h+1}{2^n}}=\\U_{\frac{2(h+1)}{2^{n+1}}}=U_{\frac{(2h+1)+1}{2^{n+1}}}=U_{\frac{k+1}{2^{n+1}}} $$ where we used $\eqref{eq:simple1}$ and the inductive hypotesis: thus we conclude that the above inclusion holds for $n+1$ and so finally by induction for any $n\in\Bbb N$.
Now Munkres says that we must proceed as in Urysohn lemma so that let's we assume $\eqref{eq:simple2}$ implies that for any $\delta_1,\delta_2\in\Delta$ such that $\delta_1\le\delta_2$ the inclusion $$ \begin{equation}\tag{3}\label{eq:simple3}\operatorname{cl}U_{\delta_1}\subseteq U_{\delta_2}\end{equation} $$ holds so that let's we prove that the position $$ \begin{equation}\tag{4}\label{eq:simple4}f(x):=\inf\{\delta\in\Delta:x\in U_\delta\}\end{equation} $$ defines a continuous function $f$ from $X$ to $[0,1]$ such that $$ f(e)=0\quad\text{and}\quad f[C]\subseteq 1 $$
So by Archimedean property we observe that for any $\delta\in\Delta$ there exists $n\in\Bbb N$ such that $$ \frac 1 n<\delta $$ so that by $\eqref{eq:simple3}$ we conclude that $$ e\in V_{n}=U_{\frac 1 n}\subseteq U_\delta $$ which proves that $$ f(e)=0 $$ Moreover by $\eqref{eq:simple1}$ and $\eqref{eq:simple3}$ we observe that for any $\delta\in\Delta$ with $\delta\le 1$ the inclusion $$ U_\delta\subseteq U_1=V_0\subseteq X\setminus C $$ holds so that also the identity $$ f[C]\subseteq\{1\} $$ holds.
So we have only to prove continuity showing that $f^{-1}\big[[0,a)\big]$ and $f^{-1}\big[(a,1]\big]$ are open for any $a\in(0,1)$.
First of all for any $x\in X$ we let put $$ \Delta_x:=\{\delta\in\Delta:x\in U_\delta\} $$
Now let be $$ U_-:=\bigcup_{\delta<a}U_\delta $$ and so we observe that if $x\in U_-$ then $x\in U_\delta$ with $\delta<a$ which implies that $$ \inf \Delta_x\le \delta<a $$ so that $x\in f^{-1}\big[[0,a)\big]$; conversely if $x\in f^{-1}\big[[0,a)\big]$ then $\inf\Delta_x<a$ so that there exists $\delta_a\in\Delta_x$ such that $x\in U_\delta$ with $\delta_a<a$ and so $x\in U_-$: thus we conclude that $$ f^{-1}\big[[0,a)\big]=U_- $$ which proves that $f^{-1}\big[[0,a)\big]$ is open.
Now let be $$ U_+:=\bigcup_{\delta>a}X\setminus\operatorname{cl}U_\delta $$ and so we observe that if $x\in f^{-1}\big[(a,1]\big]$ then $x\in U_+$ because if $x\notin U_1$ then $\eqref{eq:simple3}$ implies that $x\notin\operatorname{cl}U_\delta$ for any $\delta\in\Delta$ with $\delta<1$ and because if $x\in U_1$ then the inequality $$ \inf\Delta_x=f(x)>a $$ implies that $x\not\in\operatorname{cl}U_\delta$ for any $\delta\in\Delta$ with $a<\delta<f(x)$ since $\Delta$ is dense so that there exists $d\in\Delta$ such that $$ a<\delta<d<f(x) $$ and so by $\eqref{eq:simple3}$ if $x\in\operatorname{cl}U_\delta$ then $x\in U_d$ and this would disagree with the inequality $$ d<\inf\Delta_x $$ on the contrary if $x\in U_+$ then $x\in X\setminus\operatorname{cl}U_\delta$ for $\delta>a$ but if $d\in\Delta$ is less than $\delta$ then by $\eqref{eq:simple3}$ the inclusion $$ \operatorname{cl}U_d\subseteq U_\delta\subseteq\operatorname{cl}U_\delta $$ holds and so it must also be $$ X\setminus\operatorname{cl}U_\delta\subseteq X\setminus\operatorname{cl}U_d $$ so that necessarily $x\notin U_d$ for any $d\in\Delta$ with $d\le\delta$ and so if $x\in U_d$ then necessarily $\delta<d$ so that $$ a<\delta\le\inf\Delta_x=f(x) $$ that is $x\in f^{-1}\big[(a,1]\big]$: so we finally conclude that $$ f^{-1}\big[(a,1]\big]=U_+ $$ which proves that $f^{-1}\big[(a,1]\big]$ is open.
So as you can see I was not able to prove the inclusion $\eqref{eq:simple3}$ so that I ask to do it and moreover I ask if I well proved the marked parts with a grey bar since Munkres did not prove them. So could someone help me, please?
Let $\delta_1 < \delta_2$. Choose $n$ such that $\delta_1 + \frac{1}{2^n} < \delta_2$. Then from (2) : $$ V_n* U(\delta_1) \subseteq U(\delta_2).$$ Let $x$ be in the closure of $U(\delta_1)$. Since $V_n^{-1} * x$ is a neightbourhood of $x$, there exists $v \in V_n$ such that $y:= v^{-1}*x \in U(\delta_1) \cap (V_n^{-1} * x)$. You can write $$x = v *y \subseteq V_n* U(\delta_1) \subseteq U(\delta_2).$$ Hence $\mathrm{cl}(U(\delta_1)) \subset U(\delta_2)$.