Embedding of countable linear orders into $\Bbb Q$ as topological spaces
I believe the answer is yes.
Call a subset $A\subseteq \mathbf Q$ well-embedded if the intrinsic and subspace topologies coincide. The basic idea is that the failure of well-embededness is caused by a set of "holes" between elements of $A$ and monotone sequences convergent to them in $A$, and we can upgrade any subset of $\mathbf Q$ to a well-embedded one by "collapsing" the corresponding holes.
Note that the topologies coincide exactly when every point has the same neighbourhoods in both topologies. It is not hard to see that $A$ is well-embedded exactly when for each $a\in A$, if $a$ is neither minimal nor a successor (in $A$), then for every $q<a$ there is some $a'\in A$ such that $q\leq a'<a$, and if it is neither a predecessor nor maximal, then for every $q>a$, there is some $a'$ with $a<a'\leq q$.
Conversely, $A$ is not well-embedded exactly when there is some half-open interval $(a,q]$ such that $a$ is not a predecessor and not maximal in $A$ and $(a,q]\cap A$ is empty, or some half-open interval $[q,a)$ with the corresponding property. Let us call such an interval a bad interval of $A$. Note that every element of a bad interval is also the (closed) endpoint of a bad interval.
Fix any $A\subseteq\mathbf Q$. Write $L=\mathbf Q\setminus \bigcup_I I$, where $I$ ranges over bad intervals. Note that $A\subseteq L$. I claim that $L$ is dense, and so $L'=\mathbf Q+(\mathbf Q\setminus \bigcup_I I)+\mathbf Q$ is dense without endpoints.
Indeed, suppose towards contradiction that $q_1< q_2\in L$ are such that $(q_1,q_2)\cap L$ is empty. Take some $q\in (q_1,q_2)$. Then $q$ is the closed endpoint of a bad interval $(a,q]$ or $[q,a)$. Suppose the former holds (the other case is analogous). Then we cannot have $a<q_1$ (because then $q_1\in (a,q]$) and we cannot have $a>q_1$ (because then $a\in L$ and $q_1<a<q<q_2$, so $a\in L\cap (q_1,q_2)$), so we have $a=q_1$. Since $(q_1,q_2)\cap A=\emptyset$ and $a=q_1$ is not a predecessor in $A$, $q_2\notin A$. But then $(q_1,q_2]$ is a bad interval, a contradiction.
Note that $L'\cong \mathbf Q$ (because both are countable dense without endpoints). I claim that $A$ is well-embedded in $L'$. Indeed, if $a\in A$ is neither minimal nor a successor and $q<a$, then there is some $q'\in L$ with $q\leq q'<a$, and since the interval $[q',a)$ (in $\mathbf Q$) is not bad, there is some $a'\in [q',a)$, whence $q\leq a'<a$. The other case to consider is analogous, and this completes the proof.
(In fact, I'm pretty sure that if you work just a little bit harder, you can show that a) $L$ already has no endpoints, and b), for any $A\subseteq \mathbf Q$, there is a weakly monotone, piecewise linear $f\colon \mathbf Q\to \mathbf Q$ which is strictly monotone on $A$ and such that $f[A]$ is well-embedded.)
For $a\in X$, say that $a_+$ is a gap if either $a$ is the greatest element of $X$ or $a$ has a successor in $X$. Similarly, say that $a_-$ is a gap if either $a$ is the least element of $X$ or $a$ has a predecessor in $X$. Now let $Y$ be the linear order obtained by adding a copy of $\mathbb{Q}$ inside every such gap of $X$ (i.e., if $a_+$ is a gap, we add a copy of $\mathbb{Q}$ immediately after $a$, and if $a_-$ is a gap, we add a copy of $\mathbb{Q}$ immediately before $a$).
I claim the inclusion $i:X\to Y$ is continuous and thus a topological embedding. Indeed, suppose $y\in Y$ and let $A=i^{-1}(y,\infty)$; we wish to show that $A$ is open in the order topology of $X$ (the case of $i^{-1}(-\infty,y)$ is similar). If $y\in X$, $A$ is simply equal to the interval $(y,\infty)$ of $X$. Otherwise, there is an element $a\in X$ such that $y$ is in the copy of $\mathbb{Q}$ in the gap $a_+$ or $a_-$. In the $a_+$ case, $A$ is equal to the interval $(a,\infty)$ of $X$. In the $a_-$ case, then $a$ has a predecessor $b$ (or is the least element in which case we set $b=-\infty$), and so $A$ is equal to the interval $(b,\infty)$ of $X$.
Finally, I claim that $Y\cong\mathbb{Q}$. Clearly $Y$ is countable, so it suffices to show it is dense and has no endpoints. For density, let $x<y$ in $Y$. If $x\not\in X$, then we can find an element between $x$ and $y$ in the copy of $\mathbb{Q}$ that $x$ is in, and similarly if $y\not\in X$. If $x,y\in X$, then either there is an element between them already in $X$, or $x_+$ and $y_-$ are both gaps, in which case $Y$ has elements added in those gaps. Similarly, if $x\in Y$ then $x$ is not a greatest or least element: this is trivial if $x\not\in X$ since then $x$ is inside a copy of $\mathbb{Q}$, and if $x\in X$ either there are larger and smaller elements already in $X$ or we added them since $x_+$ or $x_-$ was a gap.
Let me also mention that unless the topology of $X$ is discrete, it always has an order-embedding in $\mathbb{Q}$ which is not continuous. Indeed, since $X$ is not discrete, there is some $a\in X$ such that either $a_+$ or $a_-$ is not a gap; let us suppose $a_+$ is not a gap. Now take any order-embedding $f:X\to\mathbb{Q}$, and define $g:X\to\mathbb{Q}$ by $g(x)=f(x)$ if $x\leq a$ and $g(x)=f(x)+1$ if $x>a$. Then $g$ is still an order-embedding, but it is not continuous from the right at $a$.
In fact, there always exists an order-embedding $X\to\mathbb{Q}$ such that the subspace topology on the image of the embedding is discrete. You can construct such a map by iterating the process above over every non-gap of $X$, but here is a snappier description of essentially the same idea. Fix an enumeration $(x_n)$ of $X$ and define $f:X\to\mathbb{R}$ by $$f(x)=\sum_{x_n<x}\frac{1}{2^n}+\sum_{x_n\leq x}\frac{1}{2^n}.$$ Then $f$ is an order-embedding, since increasing $x$ makes each sum have more terms. However, the image of $f$ is discrete, since if $x=x_n$ then $f(y)\leq f(x)-\frac{1}{2^n}$ for all $y<x$ (from the second sum) and $f(y)\geq f(x)+\frac{1}{2^n}$ for all $y>x$ (from the first sum). Now let $Y=\mathbb{Q}\cup f(X)$; then $f$ can be restricted to an order-embedding $X\to Y$ whose image is still discrete in the order topology of $Y$ (since $Y$ is dense in $\mathbb{R}$). But $Y$ is a countable dense linear order without endpoints, so $Y\cong\mathbb{Q}$.