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.