Sum of size of distinct set of descendants $d$ distance from a node $u$, over all $u$ and $d$ is $\mathcal{O}(n\sqrt{n})$
Let us start by giving an alternative formula for $\sum_{u,d} |L(u,d)|$. For a node $x$, let $\pi^{(d)}(x)$ be its $d$'th parent, and let $h(x)$ be its height. Then
$$
 \sum_{u,d} |L(u,d)| = \sum_x |\{ L(\pi^{(d)}(x),d) : 0 \leq d \leq h(x)\}|.
$$
It suffices to show that $|\{ L(\pi^{(d)}(x),d) : 0 \leq d \leq h(x)\}| = O(\sqrt{n})$ for all nodes $x$.

Let $y = \pi^{(d-1)}(x)$ and $\pi(y)$ be $y$'s parent. If $L(\pi(y),d) \neq L(y,d-1)$ then $y$ must have a sibling $z$ with a leaf at depth $d$. We can associate the $d$ vertices on the path leading to the leaf (including $z$) to $\pi(y)$; note that vertices associated to different parents of $x$ are distinct.

For each element in $\{L(\pi^{(d)}(x),d) : 0 \leq d \leq h(x)\}$ choose a representative $\pi^{(d)}(x)$ with the maximal $d$, and let $I$ be the set of all such $d$'s. In particular, if $\pi^{(d)}(x)$ is a representative then $L(\pi^{(d)}(x),d) \neq L(\pi^{(d-1)}(x),d)$, and so $\pi^{(d)}(x)$ has $d$ vertices associated to it. We deduce that
$$
\sum_{d \in I} d \leq n.
$$
In particular, $I$ can contain at most $\sqrt{n}$ numbers larger than $\sqrt{n}$, and so $|I| = O(\sqrt{n})$.