It takes me much time to come up with this proof. Does it look fine or contain gaps?
Let $(P,<)$ and $(Q,\prec)$ be countable, dense, and linearly ordered sets without endpoints. Then $(P,<)$ and $(Q,\prec)$ are order-isomorphic.
My attempt:
Since $(P,<)$ and $(Q,\prec)$ are countable and dense, $(P,<)$ and $(Q,\prec)$ are countably infinite. Let $(p_i \mid i \in \Bbb N)$ and $(q_i \mid i \in \Bbb N)$ be their enumerations respectively.
A function $h$ from a subset of $P$ to $Q$ is call a partial isomorphism from $P$ to $Q$ if and only if $(a<b \iff h(a) \prec h(b)$ for all $a,b\in \operatorname{dom}h)$.
Claim: If $h$ is a partial isomorphism from $P$ to $Q$ such that $\operatorname{dom}h$ is finite and $p\in P, q\in Q$, then there is a partial isomorphism $h_{p,q}$ such that $p\in \operatorname{dom} h_{p,q}$ and $q\in \operatorname{ran} h_{p,q}$.
Proof: WLOG, we assume that $h=\{(p_{i_1},q_{i_1})\mid 1 \le i\le n\}$ where $p_{i_1}<p_{i_2}<p_{i_3}<\cdots<p_{i_n}$, and thus also $q_{i_1} \prec q_{i_2} \prec q_{i_3} \prec \cdots \prec q_{i_n}$.
If $p \in \operatorname{dom}h$, we are done with $p$.
If $p<p_{i_1}$, let $i_n=\min \{i\in \Bbb N\mid q_i \prec q_{i_1}\}$. The fact that $P$ has no endpoints ensures that $\{i\in \Bbb N\mid q_i \prec q_{i_1}\} \neq\emptyset$. The fact that $\Bbb N$ is well-ordered ensures that such $i_n$ exists. Then $h_p=h\cup \{(p,q_{i_n})\}$ is clearly a partial isomorphism.
If $p>p_{i_n}$, let $i_n=\min \{i\in \Bbb N\mid q_i \succ q_{i_n}\}$. Then $h_p=h\cup \{(p,q_{i_n})\}$ is clearly a partial isomorphism.
If $p_{i_k}<p<p_{i_{k+1}}$ for some $1\le k \le n$, let $i_n=\min \{i\in \Bbb N\mid q_{i_k} \prec q_i \prec q_{i_{k+1}}\}$. The fact that $P$ is dense ensures that $\{i\in \Bbb N\mid q_{i_k} \prec q_i \prec q_{i_{k+1}}\} \neq\emptyset$. The fact that $\Bbb N$ is well-ordered ensures that such $i_n$ exists. Then $h_p=h\cup \{(p,q_{i_n})\}$ is clearly a partial isomorphism.
Similarly, we can build a a partial isomorphism $h_{p,q}=h_p \cup \{(p_{i_m},q)\}$ $= h \cup \{(p,q_{i_n})\} \cup \{(p_{i_m},q)\}$. This completes the proof.
Next we construct a sequence of partial isomorphisms recursively by $$f_0=\emptyset \text{ and } f_{i+1}=(f_i)_{p_i,q_i}$$
It's clear that $(f_i\mid i\in \Bbb N)$ is an increasing sequence of partial isomorphisms with respect to $\subseteq$. Thus $\bigcup\limits_{i\in \Bbb N} f_i$ is a partial isomorphism. Moreover, $\operatorname{dom}f=P$ and $\operatorname{ran}f=Q$ by the construction of $f_i$ for each $i\in\Bbb N$. Hence $f$ is a order-isomorphism between $P$ and $Q$.
Update:
A good catch by @DanielWainfleet: $h \subsetneq h_{p,q}$.