Prove/disprove the convergence in probability (Thanks for your attention!)

The Original Problem:
Given $N,D\in\mathbb{Z}^+~(D\ge N)$, a sequence $p=\{p_i\}_{i=1}^N$ is generated by randomly selecting $N$ distinct numbers from $\{1,2,\cdots, D\}$.
Please prove or disprove the following proposition:
$$
\forall s>0,~~\mathbb{P}[\min_{i\ne j}\{|p_i-p_j|\}>s]\rightarrow 1~~(D\rightarrow \infty).
$$

Some of My Efforts:
I may know that there are $\left( \begin{array}{c}
 D\\
 N\\
\end{array} \right) $ selection cases in total. Intuitively, I guess that the minimal number distance will tend to be infinite as $D$ increases since the range goes larger and larger. I make some numerical simulations by writing a toy Python program, and get the experimental result in the case of $N=3$ as follows:

I see that the mean value of $\left(\min_{i\ne j}\{|p_i-p_j|\}\right)$ goes larger as $D$ grows. However, I have no idea about how to exactly analyse the things happened in the process of $D\rightarrow \infty$.
I am trying to find a function $f(D)$ to approximate the growing speed of the case number of $\left(\min_{i\ne j}\{|p_i-p_j|\}>s\right)$. But I am failed, since I find that the analysis of the "selection" operation is quite difficult for me.

(Some of the above statements may not be accurate or correct. Thanks again for reading such a long post!)
 A: First of all it's easier to upper boud a small probability than to lower bound a large one. So I'll consider the complement $\{\min_{i\ne j}\{|p_i-p_j|\}\leq s\}$.
If $\min_{i\ne j}\{|p_i-p_j|\}\leq s$ then at least one block of $s+1$ consecutive elements of $\{1,\ldots D\}$ contains at least two points.
A given block of $s+1$ consecutive elements of $\{1,\ldots D\}$ containing at least two points entails that one of the $N(N-1)$ pairs of two points ends up in that block, which has probability $\dfrac {(s+1)s}{ {D(D-1)}}$
And there are $D-s$ such blocks.
In the end by repeated union bound, $P(\min_{i\ne j}\{|p_i-p_j|\}\leq s) \leq (D-s) N(N-1) \dfrac {(s+1)s}{ {D(D-1)}}$.
This goes to zero as $D\to\infty$, the rest being fixed.

Let's know make this formal. My discussion above boils down to the following inclusion of events
$$
 \{\min_{i\ne j}\{|p_i-p_j|\}\leq s\}
\subset \bigcup_{1 \leq k \leq D-s} \bigcup_{1\leq i \neq j \leq N} \{p_i,p_j\in[k,\ldots,k+s]\}.
$$
So by union bound,
$$
P(\min_{i\ne j}\{|p_i-p_j|\}\leq s) \leq \sum_{1 \leq k \leq D-s} \sum_{1\leq i \neq j \leq N} P(p_i,p_j\in[k,\ldots,k+s])
$$

*

*Each summed probability is the same and equals $\dfrac {(s+1)s}{ {D(D-1)}}$: indeed $(p_i,p_j)$ is uniformly chosen among the $D(D-1)$ pairs of distinct points of $[1,D]$. Amond them, $(s+1)s$ appen to lie in the interval $[k,\ldots,k+s]$.

*The inner sum has $N(N-1)$ terms

*The outer sum has $D-s$ terms.

In the end the bound is equal to $(D-s)N(N-1)\dfrac {(s+1)s}{ {D(D-1)}}$ as claimed (and you see the multiplication just comes from repeated addition!)
