Here is Prob. 6, Sec. 24, in the book Topology by James R. Munkres, 2nd edition:
Show that if $X$ is a well-ordered set, then $X \times [0, 1)$ in the dictionary order is a linear continuum.
Here is the definition of well-ordered set, from Sec. 10 in Munkres:
A set $A$ with an order relation $<$ is said to be well-ordered if every non-empty subset of $A$ has a smallest element.
And, here is the definition of linear continuum:
A simply ordered set $L$ having more than one element is called a linear continuum if the following hold:
(1) $L$ has the least upper bound property.
(2) If $x < y$, there exists $z$ such that $x < z < y$.
My Attempt:
Of course, $X \times [0, 1)$ has more than one (in fact uncountably many) elements, even if $X$ is a singleton set.
Suppose $x_1 \times r_1$ and $x_2 \times r_2$ are any two elements of $X \times [0, 1)$ such that $x_1 \times r_1 \prec x_2 \times r_2$. Then either $x_1 \prec_X x_2$, or $x_1 = x_2$ and $r_1 < r_2$.
If $x_1 \prec_X x_2$, then the ordered pair $x_1 \times \frac{r_1+1}{2}$, for instance, is in $X \times [0, 1)$, and also $$ x_1 \times r_1 \prec x_1 \times \frac{r_1+1}{2} \prec x_2 \times r_2. $$ Let us put $$x \times r \colon= x_1 \times \frac{r_1+1}{2}.$$
On the other hand, if $x_1 = x_2$ and $r_1 < r_2$, then the ordered pair $x_1 \times \frac{r_1+r_2}{2}$, for instance, is in $X \times [0, 1)$, and also $$ x_1 \times r_1 \prec x_1 \times \frac{ r_1 + r_2 }{2} \prec x_2 \times r_2. $$ Let us put $$x \times r \colon= x_1 \times \frac{r_1+r_2}{2}.$$
In either case, we can find an element $x \times r \in X \times [0, 1)$ such that $$ x_1 \times r_1 \prec x \times r \prec x_2 \times r_2, $$ whenever $x_1 \times r_1$ and $x_2 \times r_2$ are any elements of $X \times [0, 1)$ such that $x_1 \times r_1 \prec x_2 \times r_2$.
Now let $A$ be a non-empty subset of $X \times [0, 1)$ such that $A$ is bounded from above in $X \times [0, 1)$. Then there is an ordered pair $x \times r$ in $X \times [0, 1)$ such that, for every ordered pair $y \times s \in A$, we have $$ y \times s \preceq x \times r, $$ that is, either $y \prec_X x$, or $y = x$ and $s \leq r$.
Let $\pi_1 \colon X \times [0, 1) \to X$, $z \times t \mapsto z$, be the projection map of $X \times [0, 1)$ onto $X$. Then the set $$ \pi_1 (A) \colon= \left\{ \ \pi_1 (y \times s) \ \colon \ y \times s \in A \ \right\} = \left\{ \ y \in X \ \colon \ y \times s \in A \mbox{ for some } s \in [0, 1) \ \right\} $$ is bounded from above in $X$, the element $x$ being an upper bound of this set. Thus the set of all the upper bounds in $X$ of the set $\pi_1(A)$ is a non-empty subset of the well-ordered set $X$ and so has a smallest element $x_0$, say.
If $x_0 \not\in \pi_1(A)$, then, for every element $y \times s \in A$, we have $ y \prec_X x_0$. So $x_0 \times 0$ is an upper bound of $A$ in $X$.
Moreover, if $x \times r$ is any upper bound of $A$ in $X$, then, as we have seen above, $x$ would then be an upper bound of the set $\pi_1(A)$, and so $x_0 \preceq_X x$ must hold (because $x_0$ is the least upper bound of $\pi_1(A)$). Therefore, $x_0 \times 0 \preceq x \times r$, thus showing that $x_0 \times 0$ is the least upper bound of $A$ in $X \times [0, 1)$.
On the other hand, if $x_0 \in \pi_1(A)$, then the set $$ \left( \left\{ x_0 \right\} \times [0, 1) \right) \cap A = \left\{ \ x_0 \times s \in A \ \colon \ s \in [0, 1) \ \right\} $$ is a non-empty subset of $A$, and this subset has the order type of $[0, 1)$ (i.e. there is a bijective, order-preserving mapping of this set with $[0, 1)$).
Now let $\pi_2 \colon X \times [0, 1) \to [0, 1)$, $z \times t \mapsto t$, be the projection of $X \times [0, 1)$ onto $[0, 1)$.
Then the set $$ \pi_2 \left( \left( \left\{ x_0 \right\} \times [0, 1) \right) \cap A \right) = \left\{ \ \pi_2 \left( x_0 \times s \right) \ \colon \ x_0 \times s \in A \ \right\} = \left\{ \ s \in [0, 1) \ \colon \ x_0 \times s \in A \ \right\} $$ is a non-empty subset of $[0, 1)$.
If the set $\pi_2 \left( \left( \left\{ x_0 \right\} \times [0, 1) \right) \cap A \right)$ is bounded above by some element $r$ in $[0, 1)$, then this set is also bounded above in $\mathbb{R}$ and so has a least upper bound $r_0$, say. Then $r_0 \leq r < 1$. But since $\pi_2 \left( \left( \left\{ x_0 \right\} \times [0, 1) \right) \cap A \right) $ is a non-empty subset of $[0, 1)$ which is bounded above by $r_0$, we can also conclude that $0 \leq r_0$. Therefore $r_0 \in [0, 1)$. Then $x_0 \times r_0$ is the least upper bound of $A$ in $X \times [0, 1)$.
So let us assume that the set $\pi_2 \left( \left( \left\{ x_0 \right\} \times [0, 1) \right) \cap A \right) $ is not bounded above in $[0, 1)$.
Now if $x_0$ were the largest element of $X$, then set $A$ would not be bounded from above in $X \times [0, 1)$. But since $A$ by our supposition is bounded from above in $X \times [0, 1)$, we can conclude that $x_0$ cannot be the largest element of $X$ and thus the subset $$ \left\{ \ x \in X \ \colon \ x_0 \prec_X x \ \right\}$$ of the well-ordered set $X$ must be non-empty; let $x_1$ be the smallest element of this set. Then the subset $$ \left( x_0, x_1 \right)_X = \left\{ \ x \in X \colon \ x_0 \prec_X x \prec_X x_1 \ \right\}$$ of $X$ is empty. Then $x_1 \times 0$ is the least upper bound of $A$ in $X \times [0, 1)$.
Is each and every step in the logic of my proof correct? If so, then is my presentation clear and accessible enough? If not, then where am I lacking?
I would plead for a shorter route for proving that $X\times[0,1)$ has least upperbound property.
If $A\subseteq X\times[0,1)$ is bounded from above then $$\{z\in X\mid\exists y\in[0,1)[\langle z,y\rangle\text{ is an upperbound of }A\}\neq\varnothing\tag1$$
Let $x$ be the least element of this set.
Then the set $\{s\in[0,1)\mid \langle x, s\rangle\in A\}\subseteq[0,1)$ is bounded from above hence has a least upperbound.
Denoting this upperbound by $y$ it is not difficult to prove that $\langle x,y\rangle$ is the least upperbound of $A$.
If $\langle x',y'\rangle$ is an upperbound of $A$ then $x'$ is element of the set mentioned in $(1)$ so that $x\leq x'$. If $x<x'$ then $\langle x,y\rangle<\langle x',y'\rangle$ and if $x=x'$ then $y\leq y'$ so that $\langle x,y\rangle\leq\langle x',y'\rangle$ and we are ready.