Munkres, Theorem 16.4 I don't understand the highlighted parts of the proof. 
For the first highlighted part. The statement "$(a,+\infty)\cap Y$ and $(-\infty,a)\cap Y$ form a subbasis for the subspace topology of $Y$" means that every set that is open in the subspace topology of $Y$ is a union of finite intersection of elements of the form $(a,+\infty)\cap Y$ and $(-\infty,a)\cap Y$. Any open set in the subspace topology of $Y$ is of the form $Y\cap U$ with $U$ open in the order topology of $X$. ($U$ is open in $X$ if it is a union of sets of the form $[a_0,b),[b_0,a),(a,b)$.) So the above statement says that any set $Y\cap U$ with $U$ open in $X$ is a union of a finite intersection of sets of the form $(a,+\infty)\cap Y$ and $(-\infty,a)\cap Y$. I don't see why this follows directly from what had been said before in the first paragraph of the text. And the implication that follows is unclear for now.
For the second highlighted part. We need to show that given a set open in the order topology of $Y$, it is open in the subspace topology of $Y$. The former means that the set is a union of open rays of $Y$. Each open ray of $Y$ is the intersection of $Y$ with an open ray of $X$, so each open ray of $Y$ is open in the subspace topology of $Y$. Thus the initial set is open in the subspace topology of $Y$, being a union of open sets. What for do we need to talk about subbases?

 A: It is a general fact that if $Y \subseteq X$ and $\mathcal{S}$ is a subbase for the topology of $X$, then $\mathcal{S}_Y = \{S \cap Y: S \in \mathcal{S}\}$ is a subbase for the subspace topology of $Y$. That in itself follows from the fact that if $\mathcal{S}$ is any subbase and $\mathcal{S} \subseteq \mathcal{T}'$ for some other topology $\mathcal{T}'$ on the same set then $\mathcal{T}(\mathcal{S}) \subseteq \mathcal{T}'$ by minimality.
Now assume $X$ has the order topology w.r.t. some order $<$ on $X$.
Because the open rays of the form $(a,+\infty)$ and $(-\infty,a)$ form a subbase for the order topology on $X$ by definition, the above fact explains the first sentence of the first highlighted part. So we know that all rays of $X$ intersected with $Y$ for a subbase for the subspace topology on $Y$. We also just saw in the first paragraph of the proof that $(a, +\infty) \cap Y$ is either a ray of the restricted order $<_Y$ on $Y$ or $\emptyset$ or $Y$ because $Y$ is convex. So all subbase elements of the subspace topology are $<_Y$-open and so the subspace topology is a subset of the topology induced by $<_Y$ as an order topology on $Y$.
The last highlight: a right-ray in $<_Y$ is a set 
$(a, +\infty)_Y := \{y \in Y: y >_Y a\}$ for some $a \in Y$ and this equals $(a, +\infty) \cap Y$ by definition. The same holds for the left-rays. So the $<_Y$-subbasic sets are subspace open (being $X$-open sets intersected with $Y$), and so the $<_Y$-order topology on $Y$ is a subset of the subspace topology on $Y$. Note that that part holds regardless of convexity of $Y$. 
