In Masur and Minsky's paper, p.14, Proposition 3.6 Luo's proof, it says "but $\gamma_n\to \mu$ and $\mu$ maximal implies $\beta_n \to \mu$", which I feel confused: when $\gamma_n\to \mu$ and $\beta_n$ and $\gamma_n$ are disjoint, I think $\gamma_n$ doesn't necessarily bring $\beta_n$ to converge to $\mu$. Even, thinking intuitively, it seems possible for $\beta_n$ to be farther from $\mu$ as $n$ gets larger. I don't see how combined with $\mu$ maximal would force $\gamma_n$ to make $\beta_n$ converge to $\mu$ as $\gamma_n$ converges to $\mu$.
Notations: $\mu$ is a maximal geodesic lamination and let $\gamma_i$ be any sequence of closed geodesics converging geometrically to $\mu$. For each $\gamma_n$ we may then find $\beta_n$ with $d(\beta_n, \gamma_n) = 1$ and $d(\gamma_0, \beta_n) = N-1$.
Basically you should be thinking about $\mu$ as the stable lamination coming from a pseudo-Anosov. To help a bit more with the intuition, you should think about $\gamma_n$ as a very long curves on the surface "going all over", and longer and more "all over" as $n$ increases(the sequence of curves is not contained on any subsurface). This will mean $\beta_n$ will have to follow $\gamma_n$ for a long time (so will look like $\gamma_n$, which is why we get this convergence) to be disjoint. At least that is the basic intuition you should have/develop.
Say $\beta_n \to \beta$(maybe passing to subsequence). By disjointness of $\beta_n$ and $\gamma_n$ we get $\beta$ is "disjoint" from $\mu$, so $\beta \cup \mu$ is a lamination. With the assumption that $\mu$ is maximal we get $\beta$ must be a sublamination. The unstated jump in the sketch is that $\mu$ is minimal in the sense that any leaf in the lamination is dense in the lamination(this happens in the pA case). With that you get that $\beta = \mu$. From here you can continue the proof.