Do these topologies always refine the Euclidean topology somewhere? Let $\eta$ be the usual (Euclidean) topology on $\mathbb{R}$. Say that a topology $\tau$ on $\mathbb{R}$ is almost Euclidean refining (AER) iff for every interval $I$ - allowing $\pm\infty$ as endpoints - there is some finite $F$ such that the symmetric difference $I\Delta F$ is open in $\tau$.
Trivially $\eta$ is AER and any refinement of an AER topology is again an AER. For a slightly less trivial example, fix distinct $\beta_1,\beta_2\in\mathbb{R}$. Let $$F:\mathbb{R}\rightarrow\mathbb{R}: \theta\mapsto\begin{cases}
\beta_2 & \mbox{ if $\theta=\beta_1$}\\
\beta_1 & \mbox{ if $\theta=\beta_2$}\\
\theta & \mbox{otherwise}\\
\end{cases}$$ and let $\tau=\{U: F^{-1}(U)\in\eta\}$. Then $\tau$ is AER but does not refine $\eta$. We can also get AER topologies strictly coarser than $\eta$, e.g. $\{X\in\eta: 17\in X\implies X=\mathbb{R}\}$.
However, every AER topology I can think of does "frequently" refine $\eta$. I'm curious whether this is necessary:

Suppose $\tau$ is an AER topology. Is there some nonempty open-in-the-usual-sense $U\in\eta$ such that the subspace topology on $U$ coming from $\tau$ refines (not necessarily strictly) that coming from $\eta$?

 A: Yes, there is always such a $U.$
I'll use the following principle several times. Given a non-empty collection of "troublesome" bounded open intervals $(p,q),$ either:

*

*good case: there is an open interval $(a,b)$ such that the containment $a<p<q<b$ does not hold for any troublesome interval $(p,q)$

*bad case: there is a real $x$ such that there is a sequence of troublesome $(p_n,q_n)$ with $p_n<p_{n+1}<x<q_{n+1}<q_n$ for all $n$
If we can rule out the second case, then we can restrict to the good interval and its $\tau$ subspace topology, and by applying an increasing bijection $\mathbb R\cong (a,b)$ reduce to the case that there are no troublesome intervals.
The dichotomy holds because if the partial order defined by $(a,b)\subset\subset (p,q) \iff p<a<b<q$ has no least element, it has an infinite descending chain $(p_n,q_n),$ and we can set $x=\sup p_n.$
Step 1a: Take $(p,q)$ to be troublesome whenever $q$ is in the $\tau$-closure of $\{p\}.$ In the bad case, we use the fact that there is a finite $F$ such that $(-\infty,x]\triangle F$ is $\tau$-closed. But $(-\infty,x]\triangle F$ contains cofinitely many $p_n,$ which means it contains cofinitely many $q_n,$ which is a contradiction because they would have to lie in $F.$
Step 1b: The same but with the order reversed: $(p,q)$ is troublesome if $p$ is in the closure of $\{q\}.$
Steps 1a and 1b reduce to the case that all points are $\tau$-closed (the "T1" separation axiom).
Step 2a: $(p,q)$ is troublesome if $q$ is in the $\tau$-closure of $(-\infty,p].$ In the bad case, we again use the fact that there is a finite $F$ such that $(-\infty,x]\triangle F$ is $\tau$-closed. Since points are closed, we can add them in and so assume $F\subset (x,\infty).$  But $(-\infty,x]\triangle F$ is then a superset of each $(-\infty,p_n],$ which means it contains all the $q_n,$ which again is a contradiction because these would have to lie in $F.$
Step 2b: same as 2a but with the order reversed: $(p,q)$ is troublesome if $p$ is in the $\tau$-closure of $[q,\infty).$
Steps 2a and 2b reduce to the case that $(-\infty,q]$ is $\tau$-closed for each real $q.$ This implies $\eta\subseteq\tau.$
