TITLE: Is there a subset of the plane that meets every line in two open intervals?
QUESTION [30 upvotes]: Using the Axiom of Choice, it is possible to construct a subset of the plane that meets every line in two points (these are called "$2$-point sets"). What if, instead of points, we ask for two open intervals?
Some related ideas/constructions were explored in this paper (for example, it is shown that there is a subset of the plane meeting every line in a copy of the Cantor set).
If you want one open interval instead of two, then $\mathbb{R}^2$ itself suffices (and this is the only solution). If you want three open intervals, then just take the complement of a "$2$-point set" (and this idea can be modified to give you $n$ open intervals for every $n > 2$). Infinitely many open intervals is even easier (take the union of the interiors of the hexagons in this tiling, for example). The case of two intervals seems more stubborn.
UPDATE: This question has now been answered: there is no such subset of the plane.
Terry Tao posted some "partial progress" on the problem, which later turned out to be the first half of a complete proof. The second half can be found in my answer below. Neither post stands alone as a complete answer, but both taken together do the job.
I can't accept both answers, so I've accepted Terry's, since it came first and ended up being the first half of a correct proof. This seems to me to be in line with what the help center says about what it means to accept an answer.

REPLY [13 votes]: The answer to the question is no.
Rather than typing a new answer, I've just edited my old one (which contained some partial progress). I think this is OK since the main idea of the old answer (or at least its main "trick" -- fixing a special line, and then rotating it slightly about a point not in $E$) is still present in this one. The proof I'm about to give combines this trick with the results from Terry Tao's answer.

Theorem: There is no subset of the plane meeting every line in two open intervals.

Proof: Suppose $E$ is such a set.
Following the terminology in Terry Tao's answer, we will say that $\omega \in S^1$ is a limit direction of $E$ if $E$ does not avoid an infinite open sector containing the direction $\omega$ in the limit. Terry Tao proves:
Lemma: Not every direction is a limit direction.
(Actually he proves more, but this is all we will need.)
In other words, there is an infinite open sector $S$ that misses $E$. Translating and rotating $E$ if necessary, we may assume that $S$ contains the origin $O$ and the positive $X$-axis. Consider the expression (using polar coordinates)
$$(*) \ \ \ \ \ \ \ \ \ \ \ \{(r,\theta) : 0 < \theta < \phi, r \geq 0\} \cap E = \emptyset.$$
Clearly $\pi$ does not satisfy $(*)$ (otherwise $E$ does not meet the line $y = 1$). Let
$$\phi_0 = \inf \{\phi \in [0,\pi] : \phi \text{ does not satisfy }(*)\}.$$
(Since $S$ is open, we will have $0 < \phi_0 < \pi$, although we won't actually need this fact.) Let $L$ be the line through the origin in the direction $\phi_0$.
$E$ is open (this observation is also due to Terry Tao; see his first comment on Igor Rivin's answer). It follows that the ray $\{(r,\phi_0) : r \geq 0\}$ contains no points of $E$ (if it did, then a slightly smaller value of $\phi_0$ would also fail to satisfy $(*)$). Therefore $E$ meets $L$ in two open intervals that are both on the same side of the origin $O$.
Let $U$ and $V$ be these two open intervals, labeled in such a way that $V$ is between $U$ and $O$. Fix a point $x \in L \setminus E$ with $x$ strictly between $U$ and $V$. For $\rho \in [0,\pi)$, let $R_\rho^x$ denote the rotation of the plane about the point $x$ by the angle $\rho$. Fix $u \in U$ and $v \in V$.
We have $u,v \in E$ and $O \in S$, with $E$ and $S$ both open. By the continuity (in $\rho$) of $R^x_\rho$, there is some $\rho_0$ small enough that if $0 < \rho < \rho_0$ then $R_\rho^x(u), R_\rho^x(v) \in E$ and $R_\rho^x(O) \in S$. We may assume $\rho_0 < \frac{\pi}{2}$.
By the maximality of $\phi_0$, there is some $\varepsilon < \rho_0$ such that
$$\{(r,\phi_0+\varepsilon) : r \geq 0\} \cap E \neq \emptyset.$$
In other words, we may choose a point $e \in E$ such that, if we let $p = (r,\phi_0)$ for some $r > 0$, the angle $pOe$ is less than $\rho_0$. Since angle $pOe$ is acute (recall $\rho_0 < \frac{\pi}{2}$), and since $x$ is on the other side of $O$ from $p$, the angle $pxe$ is even smaller than the angle $pOe$. 
Let $\rho$ be the measure of the angle $pxe$. The line $L' = R^x_\rho(L)$ does not meet $E$ in two open intervals. To see this, observe that $L'$ passes through $E$ near $u$, through $x$ (not in $E$), through $E$ again near $v$, through $S$ near $O$ (not in $E$), and then through $e$ (in $E$). QED