Let $(A,<)$ and $(C,\prec)$ be countable, dense, and linearly ordered sets without endpoints. Then $(A,<)$ and $(C,\prec)$ are order-isomorphic.
On the basis of @DanielWainfleet, I present a proof here. Since it's quite lengthy, it will be great and helpful if someone can help me check it out and figure out logical gaps as well as flaws. Thank you for your help!
Let $(a_n \mid n\in \Bbb N)$ be an enumeration of $A$.
Lemma 1: Let $(A,<)$ be countable, dense, and linearly ordered set without endpoints. Then there exists $B_1 \subseteq A$ such that $B_1$ is order isomorphic to $\Bbb N$ and is unbounded from above in $A$.
Proof:
We define a mapping $f:\Bbb N \to \Bbb N$ recursively by $$f(0)=0 \text{ and }f(n+1)=\min \{i\in\Bbb N\mid a_{f(n)}<a_i\}$$
We define a mapping $g:\Bbb N \to A$ by $$g(n)=a_{f(n)}$$
It follows from the definition of $f$ that $\forall n\in\Bbb N:a_{f(n)}<a_{f(n+1)}$ and thus $g$ is injective. Let $B_1:=g[\Bbb N]$.
$a_0=g(0)<g(1) \implies B_1$ is not bounded above by $a_0$. Assume that $B_1$ is not bounded above by $a_i$ for all $i\le n$. Then $\exists n_0\in \Bbb N,\forall i\le n:a_i \le a_{f(n_0)}$.
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If $a_{n+1}\le a_{f(n_0)}=g(n_0)$: $B_1$ is not bounded above by $a_{n+1}$.
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If $a_{n+1} > a_{f(n_0)}$: We have $\forall i\le n:a_i\le a_{f(n_0)}$ and $a_{f(n_0)}<a_{n+1} \implies$ $\min \{i\in\Bbb N\mid a_{f(n_0)}<a_i\}=n+1 \implies f(n_0+1)=n+1 \implies$ $a_{f(n_0+1)} = a_{n+1} \implies g(n_0+1)=a_{n+1}$. Hence $B_1$ is not bounded above by $a_{n+1}$.
Lemma 2: Let $(A,<)$ be countable, dense, and linearly ordered set without endpoints. Then there exists $B \subseteq A$ such that $B$ is order isomorphic to $\Bbb Z$ and is unbounded from above and below in $A$.
Proof:
By Lemma 1, there exists $B_1 \subseteq A$ such that $B_1$ is order isomorphic to $\Bbb N$ and is unbounded from above in $A$.
We define a reverse-order $<^*$ on $A$ by $\forall x,y\in A:x <^* y \iff y<x$. Then $(A,<^*)$ is a countable, dense, and linearly ordered set without endpoints.
By Lemma 1, there exists $B_2 \subseteq A$ such that $B_2$ is order isomorphic to $\Bbb N$ and is unbounded from above in $A$ with respect to $<^*$. Thus $B_2$ is unbounded from below in $A$ with respect to $<$.
Let $B=B_1 \cup B_2$.
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$B_1 \cap B_2 =a_0$
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$\forall x\in B_1\setminus\{a_0\},y\in B_2:y<x$
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$B_1$ is isomorphic to the set of non-negative integers and is unbounded from above in $A$ with respect to $<$.
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$B_2$ is isomorphic to the set of non-positive integers and is unbounded from below in $A$ with respect to $<$.
Hence $B$ is isomorphic to $\Bbb Z$.
Lemma 3: Let $(A,<)$ be countable, dense, and linearly ordered set without endpoints. Then there exists $\{I_k:k\in \Bbb N\}$ such that
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$a_0\in I_0.$
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$\forall k\in \Bbb N$: $I_k\subseteq A$, $I_k\subsetneq I_{k+1}$, and $I_k$ is order-isomorphic to $\Bbb Z$ and is unbounded above and below in $A$.
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If $x,y$ are consecutive members of $I_k$ with $x<y,$ there is exactly one $z \in I_{k+1}$ \ $I_k$ such that $x<z<y$ (Note: $x,y$ are consecutive members of $I_k$ if and only if no member of $I_k$ is between $x$ and $y$).
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$\bigcup_{k\in \Bbb N} I_k=A$
Proof:
By Lemma 2, there exists $B \subseteq A$ such that $B$ is order isomorphic to $\Bbb Z$ and is unbounded from above and from below in $A$.
Let $h_k$ be an order isomorphism between $\Bbb Z$ and $I_k$.
We define a mapping $h_{k+1}:\Bbb Z \to A$ by $h_{k+1}(2n)=h_k(n)$ and $h_{k+1}(2n+1)=a_{i_0}$ where $i_0=\min \{i\in\Bbb N \mid h_k(n)<a_i<h_k(n+1)\}$.
Since $A$ is dense, $\{i\in\Bbb N \mid h_k(n)<a_i<h_k(n+1)\}\neq\emptyset$. Thus such $i_0$ exists.
It is easy to verify that $h_{k+1}$ is an order isomorphism between $\Bbb Z$ and $h_{k+1}[\Bbb Z]$.
We define $\{I_k:k\in \Bbb N\}$ by $I_0=B$ and $I_{k+1}=h_{k+1}[\Bbb Z]$.
It is easy to verify that $\{I_k:k\in \Bbb N\}$ satisfies the first three properties. Finally, our task is to prove $\bigcup_{k\in \Bbb N} I_k=A$.
Let $I:=\bigcup_{k\in \Bbb N} I_k$. We next prove that $\forall n\in\Bbb N:a_n\in I$ by strong induction on $n$.
It's clear that $a_0\in B=I_0$. Thus $a_0\in I$. Assume that $a_i\in I$ for all $i\le n$. Then there exists $k\in\Bbb N$ such that $a_i\in I_k$ for all $i\le n$.
- $a_{n+1} \in I_k$
Then $a_{n+1}\in I$.
- $a_{n+1} \notin I_k$
Then $h_k(p)<a_{n+1}<h_k(p+1)$ for some $p\in\Bbb Z$ since $I_k$ is unbounded from above and below in $A$.
We have $\forall i\le n:a_i\in I_k \implies \forall i\le n:i\notin \{i\in\Bbb N \mid h_k(p)<a_i<h_k(p+1)\}$ by the fact that $h_k$ is an order isomorphism between $\Bbb Z$ and $I_k$ and that there is no $z\in\Bbb Z$ such that $p<z<p+1$.
$h_k(p)<a_{n+1}<h_k(p+1) \implies n+1\in \{i\in\Bbb N \mid h_k(p)<a_i<h_k(p+1)\}$ $\implies n+1=\min \{i\in\Bbb N \mid h_k(p)<a_i<h_k(p+1)\} \implies h_{k+1}(2p+1)=a_{n+1}$ $\implies a_{n+1} \in h_{k+1}[\Bbb Z] \implies a_{n+1} \in I_{k+1} \implies a_{n+1}\in I$.
By principle of strong induction, $\forall n\in\Bbb N:a_n\in I$. Hence $A\subseteq I$. We know that $\forall k\in\Bbb N:I_k\subseteq A$, then $\bigcup_{k\in \Bbb N} I_k \subseteq A$, then $I\subseteq A$. It follows that $I=A$ and thus $\bigcup_{k\in \Bbb N} I_k=A$.
We proceed to prove our main theorem. We define a similar family $\{J_k:k\in \Bbb N\}$ for $C$. Hence $\bigcup_{k\in \Bbb N} I_k=A$ and $\bigcup_{k\in \Bbb N} J_k=C$.
Since $I_k$ and $J_k$ are each order-isomorphic to $\Bbb Z$ for all $k\in\Bbb N$, $I_k$ is order-isomorphic to $J_k$ for all $k\in\Bbb N$.
We define recursively a family $\{F_k:k\in \Bbb N\}$ of order-isomorphism between $I_k$ and $J_k$ such that $F_k\subsetneq F_{k+1}$ for all $k\in\Bbb N$ as follows.
$F_0=T$ where $T$ is an order isomorphism between $I_0$ and $J_0$.
$F_{k+1}(z)=F_k(z)$ for all $z\in I_k$. For each $z\in I_{k+1}\setminus I_k$, there is exactly one pair of consecutive members $x,y$ of $I_k$ such that $x<z<y$. Then we define $F_{k+1}(z)=w\in J_{k+1}\setminus J_k$ such that $F_k(x)\prec w \prec F_k(y)$. Since $F_k$ is an order isomorphism between $I_k$ and $J_k$, and $x,y$ are two consecutive members of $I_k$, then $F_k(x),F_k(y)$ are two consecutive members of $J_k$. Then there exists a unique $w\in J_{k+1}\setminus J_k$ such that $F_k(x)\prec w \prec F_k(y)$. Thus $\forall z\in I_{k+1}\setminus I_k:F_{k+1}(z)$ is well-defined.
It is easy to verify that $F_k$ is an order isomorphism between $I_k$ and $J_k$ for all $k\in\Bbb N$.
Let $F=\bigcup_{k\in\Bbb N}F_k$. It is easy to verify that $F$ is an order isomorphism between $\bigcup_{k\in \Bbb N} I_k$ and $\bigcup_{k\in \Bbb N} J_k$. Thus $F$ is an order isomorphism between $A$ and $C$. Hence $A$ and $C$ are order-isomorphic.
Best Answer
A shorter, hands on approach is to show $A$ is
order isomorphic to $\Bbb Z[1/2]$, the dyadic rationals.
From that comes the fact that any $A$ is order isomorphic to $\Bbb Q$.
The proposition is a result of showing any countable linear order order embeds into the dyadic rationals. Construction of the embedding is done one element at a time by induction over a listing of $A$.