Is there a subset of the plane that meets every line in two open intervals? 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.
 A: It is a result of Frantz
\bib{MR1141290}{article}{
   author={Frantz, Marc},
   title={On Sierpi\'nski's nonmeasurable set},
   journal={Fund. Math.},
   volume={139},
   date={1991},
   number={1},
   pages={17--22},
   issn={0016-2736},
   review={\MR{1141290 (93a:28006)}},
}

That such a set (actually, Frantz allows exceptional intersections, so that his result is clearly "nonempty", for example the union of two open infinite strips is fine) is measurable. I would guess that this means that such a set does not exist.
A: Let $E$ be a set of the claimed form.  Call a direction $\omega \in S^1$ a limit direction of $E$ if there exists a sequence $p_n$ of points in $E$ going to infinity whose argument goes to $\omega$, or equivalently if $E$ does not avoid an infinite open sector containing the direction $\omega$ in the limit.  I can show the following:

Proposition.  The set of limit directions is a closed non-empty subset of $S^1$ which is not all of $S^1$.  Furthermore, if $\omega$ is a limit direction, then every line parallel to $\omega$ meets $E$ in the union of two open intervals, one of which is half-infinite in the direction $\omega$.

This doesn't settle the question yet, but may be useful partial progress towards a complete solution.  One consequence of this proposition is that if one line meets $E$ with a half-infinite interval (plus another interval), then all parallel lines do also (since the direction of the half-infinite interval is clearly a limit direction).
Proof: It is clear that the set of limit directions is closed.  Since $E$ meets every line at least once, it is unbounded.  Thus there is a sequence of points $p_n$ in $E$ going to infinity.  By Bolzano-Weierstrass this shows that there is at least one limit direction.
Suppose that the direction $(1,0)$ was a limit direction, thus we have a sequence $p_n = (x_n,y_n)$ in $E$ where $x_n \to +\infty$ and $y_n/x_n \to 0$.  By reflection and passing to subsequence we may assume that the $y_n$ are all nonnegative; by shifting $E$ upwards slightly we may assume they are all strictly positive.
By hypothesis, $E$ meets the $x$-axis in two disjoint intervals $(a,b) \times \{0\}$ and $(c,d) \times \{0\}$ with $a < b < c < d$.  Suppose that $d$ was finite.  If we choose $a < x < b < y < c < z < d < w$, then there are vertical open intervals $I_x, I_z$ around $(x,0)$ and $(z,0)$ that respectively lie in $E$, while $(y,0)$ and $(w,0)$ do not lie in $E$.  
Consider the vertical line through $(w,0)$.  This meets $E$ in two open intervals, neither of which contains $(w,0)$.  Thus there is an open interval $J_w = \{w\} \times (0,\varepsilon)$ that either lies completely outside of $E$, or completely inside $E$.  But for $n$ large enough, we can find a line that meets $I_x$, $(y,0)$, $I_z$, $J_w$, and $p_n$ in that order.  Since this line has to meet $E$ in two intervals, this forces $J_w$ to lie completely inside $E$.
We conclude: if $d$ is finite, then for sufficiently large $w$, there exists $\varepsilon>0$ such that $(w,0)$ lies outside of $E$ but $\{w\} \times (0,\varepsilon)$ lies in $E$.
Of course, for $d$ infinite, we have $(w,0) \in E$ for all sufficiently large $w$.
Shifting $E$ upwards (which keeps $y_n$ positive and $y_n/x_n$ going to zero), we conclude that for any $t \leq 0$, either $(w,t) \in E$ for all sufficiently large $w$, or else for sufficiently large $w$ (depending on $t$), there exists $\varepsilon>0$ such that $(w,t)$ lies outside of $E$ but $\{w\} \times (t,t+\varepsilon)$ lies in $E$.
If the second option holds true for at least three values of $t \leq 0$, then we conclude for sufficiently large $w$ that the indicator of $E$ on the vertical line $\{w\} \times {\mathbf R}$ changes value at least five times, and so $E$ does not meet this line in two intervals, a contradiction.  Thus the first option must hold for at least one $t \leq 0$.  In particular we now have a sequence $(x_n,y_n)$ of points with $y_n = t < 0$ and $y_n/x_n \to 0$, so by reflection all the results we had for $E$ also hold for the reflection of $E$ across the $x$ axis.  In particular, if $d$ is finite, it is now true that for sufficiently large $w$, there is an interval $K_w = \{w\} \times (-\varepsilon,\varepsilon)$ such that $K_w$ meets $E$ at every point of $K_w$ except for the midpoint $(w,0)$.  
Now by intersecting $E$ with ${\mathbf R} \times \{1\}$ we may find $f < g$ such that $(f,1) \in E$ and $(g,1) \not \in E$.  Using the vertical line $\{f\} \times {\mathbf R}$ we may find a vertical interval $L_f$ around $(f,1)$ that lies in $E$.  But one can then find arbitrarily large $w_1 < w_2 < w_3$ close together such that there is a line passing through $L_f$, $(g,1)$, $K_{w_1}$, $(w_2,0)$, $K_{w_3}$ in that order, contradicting the fact that $E$ has to meet this line in two intervals.  (To find $w_1,w_2,w_3$, one can for instance use the Lebesgue differentiation theorem to locate an arbitrarily large real $w_0$ where the length of $K_w$ is bounded from below for set of $w$ of asymptotic density $1$ near $w_0$, then set $w_1,w_2,w_3$ to be sufficiently close generic points near $w_0$.)  We conclude that $d$ must be infinite.  Translating this up and down, we now conclude that $E$ meets every horizontal line in two intervals, one of which is half-infinite to the right.
Finally one has to show that not every direction is a limit direction.  This is an observation (now deleted) of Robert Israel: if every direction was a limit direction, then every line meets the complement of E in a closed interval, so the complement of E is convex.  But by the Hahn-Banach separation theorem we can then find a line that separates a point of E from its complement, so E meets all of that line rather than meeting it in two intervals, a contradiction. $\Box$
One can say a bit more about the limit directions.  If $\omega_1,\omega_2,\omega_3,\omega_4,\omega_5$ lie in a semicircle in that order, one cannot have $\omega_1,\omega_3,\omega_5$ a limit direction and $\omega_2,\omega_4$ not, since in that case there would be rays in the directions $\omega_3,\omega_5$ that were in $E$ and rays in the directions $\omega_2,\omega_4$ that lay outside of $E$, which contradicts the proposition in the direction $\omega_1$.  I think this means that the set of non-limit directions consists of at most three open intervals.
A: 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
