Open rays form a subbasis for the order topology on $X$

Claim: If $X$ is a ordered set, equipped with $\mathcal{T}$ order topology, then set $S=\{ (a,+\infty)|a\in X\} \cup \{(-\infty,a)| a\in X\}$ of open rays of $X$ is a subbasis for order topology on $X$.

I find, proof of this claim unsatisfactory in Munkres topology book. This lemma appears in the last part of section 14(order topology).

My attempt: let $A=\{\bigcap_{i\in J_{n}} S_i|S_i \in S, n\in \Bbb{N}\}$ be a basis generated from $S$. $\mathcal{B}=\{ (a,b)|a,b\in X, b\gt a\} \cup \{[a_0, b):a_0$ is smallest element of $X$, $b\gt a_0\} \cup \{ (a,b_0]:b_0$ is greatest element of $X, b_0\gt a\}$ is basis for $\mathcal{T}$. We’ll prove this claim using lemma 13.2. It is easy to check $S\subseteq \mathcal{T}$. Let $\cap_{i\in J_{n}} S_i=M\in A$. Since $S_i \in \mathcal{T}$(order topology) $\forall i \in J_n$, $\cap_{i\in J_{n}} S_i \in \mathcal{T}$. Thus $M\in \mathcal{T}$. Hence $A\subseteq \mathcal{T}$.

Now we show $\mathcal{B}\subseteq A$. If $(a,b)\in \mathcal{B}$, then $(a,+\infty)\cap (-\infty, b)=(a,b)$, since $a\lt b$. So $(a,b)\in A$. If $[a_0,b), (a,b_0] \in \mathcal{B}$, then $[a_0,b), (a,b_0]$ are open rays of $a$. So they are in $\mathcal{T}$. Let $U\in \mathcal{T}$ and $x\in U$. Then $\exists B\in \mathcal{B} \subseteq A$ such that $x\in B\subseteq U$. Thus for $x\in U, \exists B\in A$ such that $x\in B\subseteq U$. By lemma 13.2, set $A$ is the basis for order topology $\mathcal{T}$. Hence, topology generated by subbasis $S$ is equal to order topology $\mathcal{T}$. Is this proof correct?

First of all, at my old university (where many PhD-theses on ordered spaces (LOTS) have been written), in the first course in general topology the order topology was defined to be the unique topology on $(X,<)$ generated by the subbase of open rays. So the question is moot in that context. Munkres has created a bit of a fuzz (IMO) to define the topology by a base instead: it's less elegant because the base depends on whether $\min(X)$ and/or $\max(X)$ exist or not, which makes that we have to consider cases depending on that in a lot of proofs on ordered spaces. This already starts in the checking that the Munkres standard base is indeed a base for some topology at all. For subbase it's trivial (in university we had no condition on a subbase at all, so there is nothing to check, and otherwise take $x < y$ in $X$ and note that $(-\infty,y) \cup (x,+\infty)=X$ so obeying Munkres' demand that a subbase must cover $X$ (if no such $x,y$ exist, $X$ is a singleton and there is a unique topology on $X$ anyway; we don't consider these trivial spaces).

It's more convenient to start from this subbase and consider directly what the generated base by it, call it $\mathcal{B}_g$, is. (we need nothing more; no lemmas from before etc. just that we have a finite intersection of subbase elements as a member of the generated base)

• If $\min(X)$ exists, all sets of the form $[\min(X),x), x \in X$ are in $\mathcal{B}_g$ as it equals $(-\infty,x)$ by definition.
• If $\max(X)$ exists, all sets of the form $(x,\max(X)], x \in X$ are in $\mathcal{B}_g$ as it equals $(x,+\infty)$ by definition. (both a single subbase element, so a "finite intersection" trivially).
• All open intervals $(x,y), x,y \in X$ are in $\mathcal{B}_g$ as it equals $(-\infty,y) \cap (x,+\infty)$.

So Munkres' base $\mathcal{B}_m \subseteq \mathcal{B}_g$. It follows that the order topology $\mathcal T_<$ (generated by $\mathcal{B}_m$) is a subset of the topology $\mathcal{T}_S$ generated by this subbase. In short $$\mathcal{T}_< \subseteq \mathcal{T}_S\tag{1}$$

On the other hand, if we have a finite intersection $B=\bigcap_{i=1}^n S_i$ of subbase elements ($n \ge 1$) define $$L = \{i \in \{1,2,\ldots,n\}\mid \exists L_i \in X: S_i = (-\infty, L_i)\}$$ the left facing subbase elements ($L=\emptyset$ is possible) and similarly $$R = \{i \in \{1,2,\ldots,n\}\mid \exists R_i \in X: S_i = (R_i, +\infty)\}$$ the right facing subbase elements. Then if $L,R \neq \emptyset$, we have $$B=(\min \{L_i\mid i \in L\},\max\{R_i\mid i \in R\})\in \mathcal{B}_m \subseteq \mathcal{T}_<$$ and if $L=\emptyset$, we have $$B=(\max\{R_i\mid i \in R\}, +\infty) \in \mathcal{T}_<$$ and if $R=\emptyset$, we have $$B=(-\infty, \min \{L_i\mid i \in L\}) \in \mathcal{T}_<$$ (Munkres mentions "nebenbei" that open rays are order open in the text) and hence in all cases $B \in \mathcal{T}_<$ and as $B$ was an arbitrary member of the generating base $\mathcal{B}_g$ for $\mathcal{T}_S$ we have that $$\mathcal{T}_S \subseteq \mathcal T_<\tag{2}$$.

$(1)$ and $(2)$ together give equalities of topologies. The generated base by the subbase is not always exactly the standard order base as defined by Munkres, because it can contain more: all open rays (which are not in Munkres' base if $\min(X)$ or $\max(X)$ do not exist) are in the generated base always. But they give no extra open sets so it doesn't matter.