Lemma: Let $A$ be a closed, weakly open set, $p$, $q$ and $r$ three distinct points, $p \notin A$, and suppose the open line segments $pq$ and $pr$ are disjoint from $A$. Then the open triangle $T$ with vertices $p$,$q$,$r$ is disjoint from $A$. Moreover, the open segment $qr$ is either disjoint from $A$ or contained in $A$.
Proof: If $T$ intersects $A$, then there is a point $y$ of $T \cap A$ that is as far as possible from the line $L$ through $qr$. $y$ must be contained in a maximal open segment $S$ contained in $A$, and since no point of $S$ can be farther from $L$ than $y$, $S$ must be parallel to $L$. But since $S$ can't intersect the open segments $pq$ and $qr$, it must have an endpoint $y' \in T$. This endpoint is also in $A$, and also maximizes the distance from $L$, but any open segment containing $y'$ and contained in $T$ must not be parallel to $L$, contradiction.
If some point $z$ of the open segment $qr$ is in $A$, it is contained in an open segment contained in $A$, but such a segment can't intersect $T$ so it must be on the line $L$. But if not all of $qr$ is in $A$, that segment has an endpoint $z' \in qr$, and then an open segment containing $z'$ and contained in $A$ must not be on $L$, contradiction. That concludes the proof of the Lemma.
Theorem: Suppose $A$ and $B$ are disjoint nonempty, closed, weakly open sets. Then there is some line $L$ disjoint from $A \cup B$, such that both $A$ and $B$ contain translates of $L$.
Proof: Along a line segment from a point of $A$ to a point of $B$ there must be an open interval disjoint from $A \cup B$ that has one endpoint in $A$ and the other in $B$. Let $p$ be a member of such an interval. Let $C_A$ be the set of points $s$ of the unit circle $C$ such that for some $t \in (0,\infty)$, $p + t s \in A$ and for all $t' \in (0,t)$, $p + t' s \notin B$. Similarly define $C_B$, interchanging $A$ and $B$. It is not hard to show that $C_A$ and $C_B$ are open and nonempty. So there must be $s_1$ and $s_2$ in $C$ that are neither in $C_A$ nor $C_B$, dividing $C$ into arcs $C_1$ and $C_2$ where $C_A$ intersects $C_1$ and $C_B$ intersects $C_2$. But by Lemma 1, neither of those arcs can subtend an angle less than $\pi$. So both arcs must subtend $\pi$, i.e. $s_2 = -s_1$, and the line $L$ through $p$ in direction $s_1$ is disjoint from $A \cup B$.
Now let $q$ be a point of $A$ such that the open interval $pq$ is disjoint from $A$, and let $L'$ be the line through $q$ parallel to $L$. Let $U$ be the open strip between $L$ and $L'$. $U$ is the union of the open triangles with vertices $p$, $q$ and points $r$ of $L$. Since the open segments $pq$ and $pr$ are disjoint from $A$, the Lemma says $U$ is disjoint from $A$. Now there is an open segment containing $q$ and contained in $A$, and since this segment can't intersect $U$ it must be contained in $L'$. Applying the second part of the Lemma to a triangle with vertices $p$, $q$ and another point of $L'$, we see that $L' \subseteq A$.
Corollary: In a partition of ${\mathbb R}^2$ into closed, weakly open sets, the boundary of each member of the partition consists of parallel straight lines.