As I understand correctly: $\#\{i=1,...,n | nD_i\geq z\}$ means the number of cases which satisfy condition $nD_i\geq z$. If so:
The limiting distribution of $n\cdot min(U_1,...,U_n)$ has an expotential distribution with parameter $\lambda = 1$. See
proof here.
Conditionally, given $U_1$, we can calculate distribution of second variable correcting for $U_1$, so the distance [0-1] was the same:
$$((n-1)\cdot min(U_2,...,U_n)-U_1)\cdot(1-U_1) | U_1$$ The result would be the same. We have $n-1$ random variables uniformly distributed at range [0-1], but in infinity $n$ and $n-1$ are the same. And this holds for every $U_1$, so we can infer, that unconditionally, no matter the randomisation, distance between first and second variable has also expotential distribution with parameter $\lambda = 1$. Now this way we can calculate recursively, that every spacing between two consecutive variables has same distribution as the minimal random variable.
If we define variables $Z_1,...,Z_n$ which are equal to $1$ if the difference would be greater or equal than $z$ and $0$ otherwise.
From expotential distribution we know that:
$$P(nD_i\geq z) = e^{-z} = E(Z_i)$$
and i guess this is it, because when limiting with $n \rightarrow \infty$, what actually is defined is expected value.
If looking for the distribution of such cases when $n$ is not going to infinity, we would need to use Central Limit Theorem. Calculate variance and expectation of $Z_i$ which is simple, and use CLT with $n$, $Var({Z_i})$ and $E(Z_i)$.
Some R codes showing the same with RNG:
#Distribution of D_i
n = 10000000
vec = diff(sort(c(runif(n),0)))*n
mean(vec)
var(vec)
hist(vec, freq=F)
#Distribution of Z_i for finite n
library(tseries)
rep = 10000
n = 100
z = 0.5
dist = c()
for(i in 1:rep){
vec = diff(sort(c(runif(n),0)))*n
dist = c(dist,sum(vec>z))
}
dist=dist/n
mean(dist)
var(dist)^0.5
hist(dist, breaks=30)
jarque.bera.test(dist)