The Reeb foliation of the solid torus has the boundary as a leaf

(The disk, submersion and level sets.) The usual Cartesian charts cover the disk interior. Near each boundary point, choose a branch of the polar angle $\theta$ and use $(\theta ,1-r)$ as a half-space chart; its inverse is $((1-s)\cos \theta ,(1-s)\sin \theta )$ and extends smoothly to negative $s$. Overlap changes and their inverses extend smoothly, giving the disk its smooth boundary structure by F5. The map $z\mapsto z\mathrm{/}\sqrt{1-\mathrm{\mid }z{\mathrm{\mid }}^2}$ from the open disk to ${\mathbb{R}}^2$ has smooth inverse $w\mapsto w\mathrm{/}\sqrt{1+\mathrm{\mid }w{\mathrm{\mid }}^2}$, so the disk interior is diffeomorphic to the plane. On the open solid cylinder $\mathrm{Int}{\bar{D}}^2\times \mathbb{R}$ the differential of $f(x,y,t)=u(r)-t$ has ${\mathrm{\partial }}_{t}f=-1$, so $f$ is a submersion [F7]. By [F1] its kernel distribution is integrable and the maximal integral manifolds are the connected level sets, which therefore define a regular codimension-one foliation [F2]. For a fixed level $c$, the level set is the graph $\{(x,y,u(r)-c):(x,y)\in \mathrm{Int}{\bar{D}}^2\}$ of a smooth function over the disk interior, hence is diffeomorphic to $\mathrm{Int}{\bar{D}}^2\cong {\mathbb{R}}^2$; so all leaves are planes [F5, F7].

(Accumulation after the circle quotient.) Translation $t\mapsto t+1$ sends the level $c$ to the level $c-1$. The image of a level graph in $\mathrm{Int}{\bar{D}}^2\times (\mathbb{R}\mathrm{/}\mathbb{Z})$ is embedded intrinsically as a plane: its disk projection is injective and its chartwise inverse is smooth. At any boundary point with meridional angle ${\theta }_0$ and longitude $t_0<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>1$, choose large integers $k$ and the unique radii $r_k$ satisfying $u(r_k)=c+t_0+k$. Since $u$ is increasing with image $[e,\mathrm{\infty })$ and diverges at one, $r_k\to 1$; the points $(r_k,{\theta }_0,u(r_k)-c<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>1)$ on the quotient leaf tend to that boundary point. Thus every interior quotient leaf accumulates on the entire boundary torus. This conclusion is about the circle quotient: a graph in the unquotiented cylinder has $t\to +\mathrm{\infty }$ as $r\to 1$ and does not accumulate at a finite boundary-cylinder point.

(Holonomy of the boundary leaf and non-stability.) Parametrize a one-sided radial transversal near a boundary point by $r<1$; following the loop of the $S^1$-factor once returns to the same transversal at the parameter $r^{\mathrm{\prime }}$ determined by $u(r^{\mathrm{\prime }})=u(r)+1$, which exists and is unique because $u$ is strictly increasing with image $[e,\mathrm{\infty })$ and satisfies $r^{\mathrm{\prime }}>r$; as $r\to 1^{-}$ we have $r^{\mathrm{\prime }}\to 1^{-}$ [F7]. The transport germ is therefore the non-identity one-sided contraction $r\mapsto r^{\mathrm{\prime }}(r)$; its iterates $r\mapsto r_n$ with $u(r_n)=u(r)+n$ are again non-identity near the boundary, so the holonomy group of the boundary leaf is infinite. If the boundary leaf were stable, then the open neighbourhood $W=\{r>1\mathrm{/}2\}$ of the boundary leaf would contain a saturated neighbourhood $U$ of it [F8]; but $U$ is open and contains the boundary leaf, hence contains a point $p$ with $r(p)$ close to $1$, and being saturated $U$ contains the whole leaf through $p$, which is a plane meeting the circle $r=1\mathrm{/}2$ and so is not contained in $W$. This contradiction shows the compact boundary leaf with infinite holonomy is not stable, while the interior leaves are planes accumulating on it.

(Smooth boundary tangency and quotient.) Near $r=1$ set $v(r):=1\mathrm{/}{u}^{\mathrm{\prime }}(r)=(1-r^2{)}^2\exp (-1\mathrm{/}(1-r^2))\mathrm{/}(2r)$. This function extends smoothly by zero at and beyond $r=1$, with every derivative zero there: each differentiated term is a polynomial in $(1-r^2{)}^{-1}$ times the exponential and a smooth factor near one, and the exponential decays faster than every power. The form $\alpha =dr-v(r)dt$ is nowhere zero, satisfies $\alpha \wedge d\alpha =0$, and has the same kernel as $df=u^{\mathrm{\prime }}(r)dr-dt$ in the interior collar. Its zero extension gives a regular integrable distribution on a collar crossing the boundary by F9. The boundary $r=1$ is an integral hypersurface; a Frobenius chart centered there makes it a central plaque, so restricting that chart to the half-collar gives genuine boundary-tangent half-space foliation charts. These agree with the interior level-set foliation and make the connected boundary cylinder one leaf. Translation in $t$ preserves $\alpha$ and the interior foliation; it is free and properly discontinuous. To apply the boundaryless quotient supplier F4 exactly, extend the disk radius to $r<1+\epsilon$ and use the zero extension of $v$ in the added collar. There the kernel of $dr-v(r)dt$ is the product foliation by $r=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$; it agrees with the interior foliation in the original collar and is translation-invariant. The $t$-translation action on this boundaryless extension is free, and only finitely many integer translates of a compact set can meet it because its $t$-projection is bounded. F4 therefore gives its smooth foliated quotient. Restrict to the invariant closed submanifold $r\le 1$: the half-space charts already obtained give this restriction its regular boundary-tangent foliation on the solid torus, with boundary leaf $T^2$ and accumulation as proved in step 1.2 [F3, F4]. The smooth nonzero form $dt-du$ in the disk interior can also be scaled by a positive function to equal $vdt-dr$ in the collar, providing a coorientation.

The boundary $\mathrm{\partial }X\cong T^2$ is a compact leaf, every other leaf is a plane accumulating on it, the foliation descends to the solid torus, and the boundary leaf has infinite holonomy and is not stable, as claimed.