Update: I have modified the answer using a more sensible definition
of $Y$. In terms of the gaps, we only care that $g_j\geq r$ for $2\leq j\leq k$
and we don't care about the size of $g_1$ or $g_{k+1}$. In this case,
the number of suitable compositions is
$n-(r-1)(k-1)\choose k$, and
$$\mathbb{P}(Y\geq r)={{n-(r-1)(k-1)\choose k}\over {n\choose k}}.$$
I have interpreted the problem as follows: you are interested in
$Y$, the smallest gap between the sampled values $X_1,\dots,X_k$.
If this isn't what you want, you may want to clarify the question.
Let's look more closely at the gaps between the $X$'s.
In addition to the gaps $G_1=X_{(1)}$ and $G_j=X_{(j)}−X_{(j−1)}$ for $2≤j≤k$,
it is convenient to introduce the final gap $G_{k+1}=(n+1)−X_{(k)}$. Here
the bracket notation refers to order statistics.
The random vector $(G_1,G_2,\dots,G_{k+1})$ is a random composition
of the number $n+1$.
That is, all outcomes $(g_1,g_2,\dots,g_{k+1})$ with
$g_1+g_2+\cdots+g_{k+1}=n+1$, $g_j≥1$
are equally likely. There are $n\choose k$ such compositions,
as found using stars and bars.
Similarly, the number
of compositions with all the $g_j\geq r$ is
$n-(r-1)(k+1)\choose k$. Therefore,
$$\mathbb{P}(Y\geq r)={{n-(r-1)(k+1)\choose k}\over {n\choose k}}.$$
From this you can work out any property of the random variable $Y$.
See this question as well:
What is the distribution of gaps?