In response to a request in the Comments from the Proposer. A proof of a result of Cantor without using a back-and-forth method. It requires much preliminary work on the structure of a linear order, followed by the proof itself, which is short. The main idea is to employ the preliminary result (III). I dk how Cantor proved this theorem..
Theorem. (Cantor). Any countably infinite linear orders $(A,<), (A',<') $ which are order-dense and without end-points are order-isomorphic.
Preliminaries. Let $<$ be a linear order on a countably infinite $A=\{a_n:n\in \Bbb N\}$ with no end-points, and order-dense in itself (i.e. if $x<y$ there exists $z$ with $x<y<z).$ Then
(I). There exists $B\subset A$ which is order-isomorphic to $\Bbb N$ and unbounded above in $A.$ Proof: Let $f(1)=1.$ Recursively, for $n\in \Bbb N$ let $f(n+1)$ be the least $j\in \Bbb N$ such that $a_{f(n)}<a_j. $ Let $B=\{a_{f(n)}:n\in \Bbb N\}.$ To show that $B$ is unbounded above in $A,$ by induction:
(I-i). $a_1=a_{f(1)}<a_{f(2)}$ so $a_1$ is not an upper bound for $B.$
(I-ii). Suppose $n\in \Bbb N$ and no member of $\{a_j:j\leq n\}$ is an upper bound for $B.$ Let $n_0$ be the least $k$ such that $a_{f(k)}\geq \max \{a_j:j\leq n\}.$ Then:
If $a_{n+1}\leq a_{f(n+0)}$ then $a_{n+1}<a_{f(n_0+1}\in B.$
If $a_{n+1}>a_{f(n_0)}$ then $n+1$ is the least $j$ such that $a_j>a_{f(n_0)}$... (because $j\leq n\implies a_j\leq a_{f(n_0)}$)... so by the recursive def'n of $f(n_0+1)$ we have $n+1=f(n_0+1).$ So $a_{n+1}=a_{f(n_0+1)}<a_{f(n_0+2)}\in B.$
(II). There exists $C\subset A$ where $C$ is order-isomorphic to $\Bbb Z$ and $C$ is unbounded above and below in $A.$ Proof: Applying (I) to the reverse-order $<^*$ on $A$ (where $x<^*y$ iff $y<x$), we obtain $B^*\subset A$ with $B^*$ order-isomorphic to the set of negative integers. So with $B$ as in (I), let $C= B\cup B^*$.
(III). $A= \{J_n:n\in \Bbb N\}$ such that
(III-i). $a_1\in J_1.$
(III-ii). Each $J_n$ is order-isomorphic to $\Bbb Z $ and is unbounded above and below in $A.$
(III-iii). $J_n\subset J_{n+1}$ for all $n.$ And whenever $x,y$ are consecutive members of $J_n$ with $x<y,$ there is exactly one $z \in J_{n+1}$ \ $J_n$ such that $x<z<y$... (Note: $x,y$ are consecutive members of $J_n$ iff no member of $J_n$ is between $x$ and $y$).
Proof: Let $J_1=\{a_1\}\cup C$ where $C$ is as in (II). Recursively define $J_{n+1}$ \ $J_n$ as follows: For any consecutive $x,y$ in $J_n$ with $x<y$ let $m$ be the least $j$ such that $x<a_j<y,$ and let $a_m$ be the unique member of $J_{n+1}$ \ $J_n$ that lies between $x$ and $y$. Then (III-i),(III-ii),(III-iii) are satisfied.
It remains to show, by induction, that the set $J=\cup_{n\in \Bbb N}J_n$ is equal to $A$. As follows, by induction:
(III-i'). $a_1\in J_1\subset J.$
(III-ii'). Suppose $\{a_j:j\leq n\}\subset J.$ Let $n_0$ be the least (or any ) $k$ such that $\{a_j:j\leq n\}\subset J_k.$
If $a_{n+1}\in J_{n_0}$ then $a_{n+1} \in J.$
If $a_{n+1}\not \in J_{n_0}$ then there are consecutive $x,y$ in $J_{n_0}$ with $x<a_{n+1}<y,$ but no member of $J_{n_0},$ and a fortiori no member of $\{a_j:j\leq n\}$ lies between $x$ and $y$, therefore $n+1$ is the least $j $ such that $x<a_j<y.$ By the recursive def'n of $J_{n_0+1}$ \ $J_{n_0},$ therefore $a_{n+1}$ is the unique member of $J_{n_0+1}$ \ $J_{n+0} $ between $x$ and $y.$ So $a_{n+1}\in J_{n_0+1}\subset J.$
$\bullet$. After all this preliminary work, to finally prove the Theorem: Let $A=\cup_{n\in \Bbb N}J_n$ and $A'=\cup_{n\in \Bbb N}J'_n$ as in (III). Since $J_1$ and $J'_1$ are each order-isomorphic to $\Bbb Z,$ let $f$ map $J_1$ order-isomorphically onto $J'_1.$
Inductively, define $f$ on each $J_n$ as follows: Suppose $f$ maps $J_n$ order-isomorphically onto $J'_n.$ If $x,y$ are any consecutive members of $J_n$ and $z$ is the unique member of $J_{n+1} \setminus J_n$ between $x$ and $y$ then let $z'$ be the unique member of $J'_{n+1} \setminus J'_n$ between $f(x)$ and $f(y).$ Now let $f(z)=z'.$ Then $f$ maps $J_{n+1}$ onto $J'_{n+1}$ order-isomorphically.
Since $J_n\subset J_{n+1}$ for all $n$ and since $A=\cup_{n\in \Bbb N}J_n,$ if $x,y\in A$ with $x<y$, then $\{x,y\}\subset J_n$ for some $n$, and $f$ is strictly order-preserving on $J_n$, so $f(x)<'f(y)$.
Remark: This theorem can be used to prove that the real interval $(0,1)$ is uncountable. Suppose not. Then there is an order-isomorphism $f:(0,1)\to \Bbb Q\cap (0,1),$ which we can extend to an order-isomorphism $f:[0,1]\to [0,1]\cap \Bbb Q$ by letting $f(0)=0$ and $f(1)=1.$ Using the fact that there is a rational beteen any two reals, we can readily show that $f$ is continuous when considered as a function from $[0,1]$ into $[0,1].$ And a basic result of analysis (the Intermediate Value Property) then implies that $[0,1]=\{f(x): 0\leq x\leq 1\}.$ But $\{f(x):0\leq x\leq 1\}=\Bbb Q \cap[0,1],$ a contradiction.
Another way is that if $(0,1)$ were countable there would be an order-isomorphism $g:(0,1)\cap \Bbb Q\to (0,1),$ but then $\{g(q):q\in \Bbb Q\cap (0,1/\sqrt 2\,)\}$ would be a non-empty subset of $(0,1)$ with no $lub .$