I have a question about the proof of Theorem 5 in Thurston's paper "A norm for the homology of 3-manifolds". It's the theorem that asserts that the fibered faces are, well, fibered. More precisely, fix a compact 3-manifold $M$, and let $B_x$ be the Thurston polytope. Then the theorem asserts that the set of elements of $H^1(M;\mathbb{R})$ which are representable by non-singular closed $1$-forms is a union of cones on open top-dimensional faces of $B_x$ (minus the origin).

Here's the part that I am having trouble with. Assume that $M$ is closed, and let $\alpha$ be any non-singular closed $1$-form. Thurston already proved that $\alpha$ lies in the cone on an open top-dimensional face of $B_x$. If $S$ is a surface in $M$, then denote by $[S] \in H^1(M;\mathbb{R})$ the Poincare dual cohomology class. Consider an incompressible surface $S$ in $M$ such that $[S]$ lies in the cone on the same open top-dimensional face of $B_x$ as $\alpha$. Most of the work of the proof goes into showing that linear combinations $t[S] + u \alpha$ with $t \geq 0$ and $u > 0$ are representable by nonsingular closed $1$-forms. That I have no problem with. However, Thurston then asserts that $[S]$ can be so represented, and the only justification he gives is that you can get it by "iterating this construction". I cannot figure out what he means here.

Can anyone help me? Thanks!