The setting is a codimension one foliation $\mathscr{F}$ of what I call $P$, the open subset of $R^n$ consisting of vectors with all positive components. I know that each leaf $L$ $\in \mathscr{F}$ is properly embedded in $P$. I also know that the normal to the tangent plane of each leaf, at every point, has all strictly positive components. Thus, there exists a nowhere vanishing vector field on $P$ which is transverse to $\mathscr{F}$, so that by Lawson $\mathscr{F}$ is transversely orientable, so has a "positive" direction. So far so good.
I would now like to place a total order on the leaves of $\mathscr{F}$ (reflexive, transitive and anti-symmetric). I understand (by a simple extension of the Jordan-Brouwer separation theorem) that each leaf $L$ separates $P$ into two disjoint open connected components, I will call them $\mathscr{O}_L^+,\mathscr{O}_L^-$, where the former is the component in the positive direction, the latter the one in the negative direction. So my first thought is, well, OK, given two leaves $A,B \in \mathscr{F}$ we will say $A \leq B$ if $B \subset \mathscr{O}^+_A$. However, I know foliations can be extremely "tricky", so I need to show for one that if $A \leq B$ we cannot also have $B \leq A$; i.e. that we can't also have $A \subset \mathscr{O}^+_B$. I could do it if I knew that there is always a transverse segment in the positive direction connecting some point of $A$ to a point of $B$, but I don't know that. I thought of using the diagonal $\Delta = P \bigcap \{x,\ldots,x)\}, x> 0$ but can't (despite some very helpful answers in another post) see how to prove $\Delta$ intersects all leaves. I've tried indexing by the distance from one leaf to another, but have found difficulties with this as well. Another approach is to say that if $A \subset \mathscr{O}^+_B$ and $B \subset \mathscr{O}^+_A$ then we would have a transversal intersecting $A$ in two points and, using the notes "Foliated Manifolds" by Haefliger, Lemma 3.3, this is impossible. However, aside from the fact that for the life of me I can't quite figure out the proof, I can't exclude the possibility that the transversal might be closed; i.e. intersect $A$ in only one point. I know I also need completeness and transitivity but my sense is if I can solve this issue then those will be an easy consequence.
I apologize for the length of this post but there was a lot of background that was unavoidable. Any ideas about where to look next would be appreciated.
Edit: Please note I do not know that either of the connected components is convex, if I did there would be no issue.
Edit: I should also mention that the idea of using transverse segments to generate an order on the leaves was taken from Novikov, but that was on a compact manifold and is not necessarily complete or anti-symmetric (though it is transitive where the definition makes sense).
Edit (response to Mike F): Let $Q \subset R^{n-1}$ consist of vectors with all positive components. Each leaf $L$ is the graph of a smooth $f: D \rightarrow R_+$, where $D$ is some open set in $Q$. Clearly, if $D = Q$ then there is no issue. But I don't know this, so if I let $\pi$ be the orthogonal projection of $P$ onto $Q$ then it comes down to proving that $D = \pi(L)$ intersects $\pi(\Delta)$. This is the sticking point. Just as a side observation, if $L$ separates $P$ then $\overline{D}$ cannot be contained in the interior of $Q$, but I don't think that helps (much).
Edit (Another approach): Not to bump the question, but I started to think (as was suggested) of using the integral curves of the normal field (which by definition are transversals). If you consider the collection of all integral curves which intersect the specific leaf $A$ (call it $\mathscr{N}_A$) then once a curve enters $\mathscr{O}^+_A$ it can't leave, since if it did it would need to intersect $A$ in a second point and that is ruled out by the result of Novikov referenced above. I just need to show somehow that $\mathscr{N}_A$ covers $\mathscr{O}^+_A$.
Edit (Key Observation): I believe Mike F has provided enough information that I should be able to prove that every ray through the origin intersects every leaf $L$ of $\mathscr{F}$. Specifically, considering the projectivization of $P$ by $R_{x > 0}$, restricted to each leaf (call it $\pi$), we turn consideration of $D$ into consideration of $X = \pi L \subset S^{n-1}$, where $S^{n-1}$ is the $(n-1)$-dimensional simplex. The key idea is: If the normals to all tangent spaces are strictly positive then it follows that every tangent space intersects a given ray $u$ in such a way that the "angle" between the space and $u$ is uniformly bounded away from zero. So we can try proving that $X$ is closed by demonstrating that, given a $u$ such that $\pi u \in \partial X$, every sequence $\{x_n \} \subset L$ such that $\pi x_n \rightarrow \pi u$ converges to a point in $u$. I haven't quite gotten this proof yet, but it's clear that the tangent space $T_{n}$ to $L$ at $x_n$ must cut $u$ at a point $z_n$ and that, given the uniformity assumptions, $z_n \rightarrow z \in u$, and then it should follow that $x_n \rightarrow z$. Thanks to Mike F.
Edit: Deleted proposed proof as it contained (embarrassing) errors. My apologies. If I come up with a complete proof then I will repost.
This should maybe be a comment, but is too long for that.
First, I wanted to note that the problem is at least somewhat subtle in the sense that the topology must come into play somehwere. If, instead of $P$, our space was $P-\Delta$ (where $\Delta$ is the diagonal ray, as in your post), then we could use helical leaves centered along $\Delta$ (still requiring positive normal vectors) to construct a foliation whose leaves are naturally in a cyclical ordering, rather than a linear ordering. Jordan-Brouwer separation would also fail here; each leaf complement has one component. The condition I suggested trying to prove in my comment, that "$P_x \cap L = \varnothing$ for $x \in L$", is also not satisfied.
Luckily, I think that the amount of topology needed is not so much. Let $X$ be the projectivization of $P$, i.e. the quotient of $P$ by the scaling action of $\mathbb{R}_{>0}$. One may concretely identify $X$ with the $(n-1)$-simplex $\{x \in P : x_1+\ldots+x_n=1\}$ if one wishes. Let $\pi : P \to X$ be the quotient map. Now, it is pretty easy to see from the assumption that $\mathscr{F}$ has positive normal vectors that, for each leaf $L$, the restriction $\pi|_L:L \to X$ is a covering map. Just show the derivative has full rank. A priori, $\pi|_L$ could fail to be surjective. However, hopefully you can confirm, using whatever argument you used to prove that the leaves were properly embedded in $P$, that $\pi$ is a proper map, so that $\pi(L)$ will be closed (as well as open).
Once you know that $\pi|_L:P\to X$ is a covering map, you can conclude that it is actually a diffeomorphism (since $X$ diffeomorphic to $\mathbb{R}^{n-1}$ and contractible). So, a teensy bit of a topology has come into play here, as we knew it must.
The fact that $\pi|_L$ is a diffeomorphism says, in effect, that $L$ intersects each of the rays $\{tv : t>0\}$ for $v \in P$, exactly once. This gives a way to linearly order your leaves (though you should check, using holonomy I suppose, that the ordering doesn't depend on the ray used).
(Added 5/Aug/2022) I think the only way forward is to prove that every leaf $L$ intersects every ray $\{tv : t >0\}$, $v \in P$. That is, $\pi(L)=X$, in the notation of my post. To do this, I think the following observation is quite important. For any fixed $u \in P$, you can check there is a positive constant $c_u$ such that, for any $v \in P$, the angle between $u$ and $v$ is in $(0,\frac{\pi}{2}-c_u)$. Since the normal vectors of $\mathcal{F}$ are in $P$, this means the angle between any tangent space of any leaf and $u$ must lie in $(c_u,\frac{\pi}{2})$. The fact that there is a uniform bound on how small the angle between $L$ and $u$ can be will, I believe, enable you to prove that $X - \pi(L)$ is open, so that $\pi(L)=X$ must hold.
Maybe the following analogous situation will make the idea clear. Suppose $S \subset \mathbb{R}^3$ is a properly embedded surface with a smoothly varying, upward-pointing (i.e. positive $z$-coordinate) normal vector. Does it follow that the projection of $S$ to the $xy$-plane is the whole $xy$-plane? No! We could take $S = \{ (x,y,\frac{1}{1-x^2-y^2}) : x^2+y^2<1\}$, whose projection is the unit disk and is asymptotic to the cylinder $x^2+y^2=1$. So, we need more hypotheses to ensure $S$ is the graph of a function $\mathbb{R}^2 \to \mathbb{R}$. Assume in addition that the tangent planes of $S$ satisfy $|\frac{\partial z}{\partial x}|,|\frac{\partial z}{\partial y}| <100$. Imposing a uniform bound on slopes like this prevents the surface from blowing up like in the latter example. To prove $S$ must project to the whole $xy$-plane, we can check the points missed are an open set. The main point here is that there is some uniform control on how much the image of a patch on $S$ can distort under the projection.