The idea that you sketched in your previous question gives you an order-preserving map of $C$ into $\Bbb R$, but the map may not be continuous: for instance, the map
$$f:[0,2]\cap\Bbb Q\to\Bbb R:x\mapsto\begin{cases}
x,&\text{if }x\le 1\\
x+1,&\text{if }x> 1
\end{cases}$$
is order-preserving but not continuous at $1$, and it could in fact be produced by your construction. What you want is possible, but one has to work a bit harder.
Let $\langle X,\preceq\rangle$ be a linear order with a countable dense subset $C$, and let $f:C\to\Bbb Q$ be an order-preserving injection. Let
$$L=\left\{x\in C:x=\sup\big(C\cap(\leftarrow,x)\big)\right\}$$
and
$$R=\left\{x\in C:x=\inf\big(C\cap(x,\to)\big)\right\}\;,$$
where the suprema and infima are taken in $X$, not $C$. Let
$$L'=\big\{x\in C:f(x)=\sup\{f(y):y\in C\cap(\leftarrow,x)\big\}$$
and
$$R'=\big\{x\in C:f(x)=\inf\{f(y):y\in C\cap(x,\to)\big\}\;.$$
Then $f$ is continuous iff $L\subseteq L'$ and $R\subseteq R'$. If $f$ is not continuous, let $B_L=L\setminus L'$ and $B_R=R\setminus R'$. If $x\in B_L$, let
$$h_L(x)=\sup\{f(y):y\in C\cap(\leftarrow,x)\}\;,$$
and if $x\in B_R$, let
$$h_R(x)=\inf\{f(y):y\in C\cap(x,\to)\}\;.$$
Define a relation $\sim$ on $\Bbb Q$ by setting $p\sim q$ iff either $p=q$, or there is an $x\in B_L$ such that $p,q\in[h_L(x),f(x)]$, or there is an $x\in B_R$ such that $p,q\in[f(x),h_R(x)]$; $\sim$ is an equivalence relation on $\Bbb Q$ with order-convex equivalence classes. Thus, $\Bbb Q/\!\!\sim$ inherits a natural linear order from $\Bbb Q$, and it’s not hard to see that it’s a dense order and hence that there is an order-isomorphism $\psi:\Bbb Q/\!\!\sim\;\to\Bbb Q$; $\psi\circ f$ is then a continuous order-isomorphism of $C$ into $\Bbb Q$. Thus, we might as well assume that the original $f$ was continuous.
You now want to extend $f$ to $X$. Unfortunately, continuity of $f$ still isn’t enough. Suppose that $X=[0,2]$, $C=\Bbb Q\cap X$, and
$$f:C\to\Bbb R:x\mapsto\begin{cases}
x,&\text{if }x<\sqrt2\\
x+1,&\text{if }x>\sqrt2\;;
\end{cases}$$
then $f$ is a continuous order-isomorphism, but because of the gap between $\sup f[C\cap(\leftarrow\sqrt2)]$ and $\inf f[C\cap(\sqrt2,\to)]$ there is no way to extend $f$ continuously to $\sqrt2$. We have to work harder yet.
Let $x\in X\setminus C$; then $x$ is not an isolated point of $X$, so either $x=\sup\big(C\cap(\leftarrow,x)\big)$ and $x$ has an immediate successor $x^+\in X$ such that $x^+\in C$, or $x=\inf\big(C\cap(x,\to)\big)$ and $x$ has an immediate predecessor $x^-\in X$ such that $x^-\in C$, or $x=\sup\big(C\cap(\leftarrow,x)\big)=\inf\big(C\cap(x,\to)\big)$.
In the first case let
$$f(x)=\sup\{f(y):y\in C\cap(\leftarrow,x)\}\;,$$
and in the second case let
$$f(x)=\inf\{f(y):y\in C\cap(x,\to)\}\;;$$
it’s not hard to check that these cases are unproblematic. In the third case let
$$\ell(x)=\sup\{f(y):y\in C\cap(\leftarrow,x)\}\quad\text{and}\quad r(x)=f(x)=\inf\{f(y):y\in C\cap(x,\to)\}\;.$$
If $\ell(x)=r(x)$ we can define $f(x)=\ell(x)=r(x)$ with no problem. The problem arises precisely when $\ell(x)<r(x)$; call such a point $x\in X\setminus C$ a bad point, and let $B=\{x\in X\setminus C:x\text{ is bad}\}$. Observe that
$$\left\{\big(\ell(x),r(x)\big):x\in B\right\}$$
is a family of pairwise disjoint non-empty open intervals in $\Bbb R$, so $B$ must be countable.
Define a relation $\approx$ on $\Bbb R$ by $s\approx t$ iff either $s=t$, or there is an $x\in B$ such that $s,t\in[\ell(x),r(x)]$; $\approx$ is an equivalence relation on $\Bbb R$ with order-convex equivalence classes, so $\Bbb R/\!\!\approx$ inherits a natural linear order from $\Bbb R$. Let $q:\Bbb R\to\Bbb R/\!\!\approx$ be the quotient map, and let $\sqsubseteq$ be the induced order on $\Bbb R/\!\!\approx$.
It’s quite straightforward to verify that $\langle\Bbb R/\!\!\approx,\sqsubseteq\rangle$ is dense, Dedekind complete, and separable and has no endpoints, so it’s order-isomorphic to $\langle\Bbb R,\le\rangle$; let $\varphi:\Bbb R/\!\!\approx\;\to\Bbb R$ be an order-isomorphism. Then $\varphi\circ f:X\to\Bbb R$ is a continuous order-embedding of $\langle X,\preceq\rangle$ into $\langle\Bbb R,\le\rangle$.
