= Solution
We use the directed set version of the <Talagrand convex distance>:
$$
d_T(x,B)=\sup_{\substack{\alpha_i\geq0\\\sum_i\alpha_i^2\leq1}}
\inf_{y\in B}\sum_{i:x_i\ne y_i}\alpha_i,\qquad
d_T(A,B)=\inf_{x\in A}d_T(x,B).
$$
Here a coordinate is an entire sampled point, so $x_i\ne y_i$ means the two planar points differ. This is the distance from the upper event $A$ to the lower event $B$; its order is relevant because the convex distance need not be symmetric.
Fix $x\in A$ and choose exactly $a$ indices $I$ whose points form a <convex chain> with the endpoints. Such a choice exists even if $L_n(x)>a$, by taking a subchain. It is a certificate of the <certifiable function> $L_n$. For any $y\in B$, let $J=\{i\in I:x_i\ne y_i\}$. The $a-|J|$ unchanged selected points remain a <convex chain> in configuration $y$, so
$$
a-|J|\leq L_n(y)\leq b,\qquad |J|\geq a-b.
$$
Choose the nonnegative weights $\alpha_i=a^{-1/2}$ for $i\in I$ and zero otherwise. They have squared sum one, and for every $y\in B$,
$$
\sum_{i:x_i\ne y_i}\alpha_i=\frac{|J|}{\sqrt a}\geq\frac{a-b}{\sqrt a}.
$$
Taking the infimum over $y$, then the supremum over weights, and finally the infimum over $x\in A$ proves
$$
\boxed{d_T(A,B)\geq\frac{a-b}{\sqrt a}.}
$$
This proof also works for $b=0$ if zero-length chains are admitted, and gives the usual infinite separation when one of the events is empty. For the uniform sampling model, $L_n\geq1$ <almost surely> when $n\geq1$.
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