Section and segment of a relation $R$ $\mathbf{Definitions:}$
$Z$ is a $R$-section of a set $X$ iff $Z\subseteq X$ and $x\in Z$ whenever $x\in X   \ \land y\in Z \land \ xRy$, $\ $for some relation $R$ on $X$
The $R$-segment of a set $X$ generated by an element $y\in X$ is $\{x\in X:xRy\}:=\text{Seg}_R(X,y)$
$\mathbf{Problem:}$
Given a set $X$ which is well ordered by a relation $R$ and $Z$ is an R-section of $X$ where $Z\neq X$. Then for some $y\in X$ we have $Z=\text{Seg}_R(X,y)$   
$\underline{\mathbf{Footnote:}}$
As is often the case, I found the answer while typing the question out here. Please feel free to see if my answer actually makes sense.
 A: IMO your proof is incomplete: you only showed that for any $x\in Z$ you have $xR y$, so that $Z\subset \text{Seg}_R(X,y)$ and inclusion in the other way is not shown. In other words, how do you know (given your proof) $xRy$ implies $x\in Z$?
For intuitive appreciation, I think it's probably better (well at least for me) to write $<$ for $R$ in this case, since $R$ is a well order anyway. Thus, interpretation of $Z$ being $R$-section is very similar to $Z$ being, say, something like $(-\infty,a)$ for some real $a$ (though reals are not well ordered!). Note, for instance, if $r\in (-\infty,a)$ and $x<r$ then it is trivially the case that $x\in (-\infty,a)$. Things like this are also called "initial segment".
Now I think there's a typo in your answer, in the second line it should say $X\setminus Z\neq \emptyset$. Then maybe the following is pretty much equivalent to your proof, but I'd go this way:
Pick least $y\in X\setminus Z$. If there is $z\in Z$ such that $y<z$ then we must have $y\in Z$ so contradiction. This shows that, since $<$ is a total order (as it is an well order), for all $z\in Z$ we have $z<y$. Thus $Z\subset \text{Seg}_R(X,y)$. Note how any element in $X\setminus Z$ would have worked for this argument, which does not require minimality of $y$
Now conversely, suppose that $x\in \text{Seg}_R(X,y)$ so that $x<y$. We'd like to show that $x\in Z$. But this is clear as follows: If $x\notin Z$, then $x\in X\setminus Z$ so that, by minimality of $y$ we must have $y<x$ or $y=x$. But since $x\in  \text{Seg}_R(X,y)$ we must also have $x<y$ by definition, and this is contradiction since $<$ must be a strict order. Thus $x\in Z$ and $\text{Seg}_R(X,y)\subset Z$ which complete proof.
A: Since $Z$ is an $R$-section of $X$, we know that $Z\subsetneq X$ since we additionally assumed $X\neq Z$.
So then $X\setminus Z\neq\emptyset$ and  $X\setminus Z\subseteq X$ so we know  $X\setminus Z$ has a $R$-least element since $X$ is well ordered by $R$.
$\require{cancel}$
So $\exists\  y\in X\setminus Z$ st. if $y\neq z \implies yRz \land\ y\cancel{R}z $ 
Since $ y\in X\setminus Z\implies y\in X \ $and$\  y\not\in Z$
Consider any $x\in Z$, then we cannot have $yRx$, since if $yRx$ then since $Z$ is an $R$-section we would have $y\in Z$, contradiction. 
So then we must have $y\cancel{R}x$.
Since $Z\subseteq X$, and since $X$ is well ordered by $R$, we also have $Z$ being well ordered by $R$. Consequently, $Z$ is linearly ordered by $R$ and hence all elements of $Z$ are comparable. 
So, $xRy$ must hold true. So we have $Z\subseteq \{x\in X: xRy\}=\ \text{Seg}_R(X,y).$
Now, let $x\in \text{Seg}_R(X,y)$. For a contradiction assume $z\not\in Z \implies z\in X\setminus Z$.
Then we must have $yRx$ since we defined $y$ to the $R$-least element. But since we assumed $x\in \text{Seg}_R(X,y)\implies xRy$
$\therefore yRx$ and $xRy \implies yRy$ by the transitive property of a well ordering. This contradicts the $($unmentioned$)$ irreflexivity property of $R$.
$\therefore x\in Z\implies \text{Seg}_R(X,y)\subseteq Z$    
$\therefore Z=\text{Seg}_R(X,y)$
