Properties of rational numbers used in Munkres proof of Urysohn's lemma (Step-$1$, Step-$2$, Step-$3$) In the first step, it involves constructing a sequence of open sets $\{U_n \}$ indexed by some re-sequencing of rational numbers $\{P_i \}$ of a topological space $X$ with the property, when $p<q$, that :
$$ U_p \subset U_q$$
(1)For constructing these sets, we arrange the set of all rational numbers in the set $\left[0,1 \right]$ into an infinite sequence in with the first two entries being $1$ and $0$. Munkres' say this is done for convenience, but I don't understand how this is so.  To clarify more, my doubt is why Munkre's went out of his way to make $1$ and $0$ the first two terms.
Now , another issue , (2)later $P_i$ is considered as consisting of the first $n$ rational number in the sequence. For the set $P_{n+1} = P_n \cup \{r \}$, it is said that $r$ has an immediate predecessor $p$ in $P_{n+1}$ and an immediate succesor $q$ in $P_{n+1}$. This defies my common sense, because, what if $r$ is larger than every other element in $P_n$ or small than? (Related)
Also, my last doubt is regarding how this proof helps us actually prove the proof of the statement of Urysohn lemma.

Urysohn's lemma: Let $X$ be a normal space, let $A$ and $B$ be disjoint closed subsets of $X$ . Let $\left[a,b \right]$ be a closed interval in the real line. Then there exists a continous map
$$ f: X \to \left[ a, b \right]$$
Such that $f(x)=a$ for every $ x \in A$ , and $f(x) = b $ for every $ x \in B$

Now here is the definition of function given in step-3:
$$ f(x) = \text{inf} Q(x) = \text{inf} \{ p | x \in U_p \}$$
(3)Now, the possible set of values for $p$ is actually rational number. So, how does the infimum help turn the function into a real valued one? Maybe this is trivial to understand if one had intuition for inf but I think some more details may help.

Btw I have linked every helpful MSE post on the Urysohn Lemma's proof based on Munkres onto this MSE post
 A: For (1): the rationals are countably infinite, as are those within $[0,1]$. Formally that means there is some given bijection $\psi:\Bbb N\to\Bbb Q\cap[0,1]$ (in fact, there are loads!). We don’t really care about the exact nature of $\psi$, but Munkres wants an easy notational access to $0,1$. As $\psi$ bijects, there are distinct and unique naturals $n,m$ with $\psi(n)=0,\psi(m)=1$. There are also, as $\psi$ injects, distinct rationals $p=\psi(1),q=\psi(2)$. Let $\varphi:\Bbb N\to\Bbb Q\cap[0,1]$ be given as follows: $$k\overset{\varphi}{\mapsto}\begin{cases}0&k=1\\1&k=2\\p&k=n\\q&k=m\\\psi(k)&\text{otherwise}\end{cases}$$You can see that $\varphi$ is a bijection. So, we can enumerate the rationals in $[0,1]$ by the sequence $\{r_n\}_{n\in\Bbb N}$, with law $r_n:=\varphi(n)$, and the first two entries of this sequence are $0,1$. In this manner, you can arrange the elements of a countable enumeration to appear in a lot of different, prescribed orders. I could, arbitrarily, demand the $123456789$th element of the sequence to be $1/(123456789)$, just for fun.
For (2): the question does not make sense to me. If $r$ is the $(n+1)$th element of the sequence, then $r$ has a predecessor in $P_n$, namely the $n$th element of the sequence, but it doesn’t have a successor in $P_{n+1}$ since the next, $(n+2)$th element would be in $P_{n+2}$... perhaps I misunderstand what you mean by successor/predecessor. Please elaborate on how Munkres uses this idea, so I can see what is actually meant.
For (3): two points. Firstly, any rational-valued function can automatically be viewed as real valued. Secondly, this function is not actually rational-valued. If I understand the context correctly, it is a continuous real function - long ago, I asked about Royden’s construction of the same function, and I will link that post if you wish. The infimum of some set of rationals might be irrational: consider $\inf_{r\in\Bbb Q}\{r>\sqrt{2}\}=\sqrt{2}$. What this function is doing (perhaps Munkres differs notably from Royden on this point, I don’t know without further details from you - but I doubt it), is getting a handle on the smallest open set $U_p$ which contains the element $x$. However, such a “smallest” $U_p$ might not exist among the collection of $U_p$, as is often the case when one deals with (continuous) infinities (what is the smallest rational $x$, $x>1$? No such $x$ exists). So, we take a limit, an infimum, and it may happen (and does happen) that the value is sometime irrational. The $U_p$ are arranged such that this limiting process produces a nice continuous function with image $[0,1]$. As a concrete example, consider the context of the Euclidean subspace $[0,1]$. If the $U_p:=[0,p)$ for rational $p\in[0,1)$, and $U_1:=[0,1]$, then what is: $$\inf_p\{1/e\in U_p\}$$It is in fact just $1/e$. Indeed, the function $f(x)$ thus constructed would end up being the identity map.
A: The convenience in (1) is that it ensures that you needn't worry about (2).
I.e. $r$ cannot be larger, or smaller, than all the elements of $P_n$ because they include 0 and 1. Without that assumption he would need to include the cases where $r$ is extremal - it's easily done but unnecessary.
