Let $S:=\mathbb X$. Let $X$ and $Y$ be iid random elements of $S$ each with distribution $\mathbb Q$. It is more transparent to rephrase the question as follows:

Can one give a good lower bound on $P(d(X,Y)\le r)$?

Let $(A_i)$ be any countable measurable partition of $S$ with each $A_i$ of diameter $\le r$. Then clearly $$P(d(X,Y)\le r)\ge\sum_i P(X\in A_i,Y\in A_i)=\sum_i P(X\in A_i)^2,$$ so that $$P(d(X,Y)\le r)\ge L(r):=\sup_{(A_i)}\sum_i P(X\in A_i)^2,$$ where the $\sup$ is taken over all countable measurable partitions of $S$ with each $A_i$ of diameter $\le r$. The lower bound $L(r)$ on $P(d(X,Y)\le r)$ is clearly attained when $S$ is an at most countable metric space with $d(x,y)>r$ for all distinct $x,y$ in $S$. One can hardly have anything better in general.

If $S=\mathbb R$ with the usual distance for $d$ and if $X$ has (say) a bounded continuous pdf $p$, then $d(X,Y)$ has the pdf $q$ given by $$q(r)=2\,1(r>0)\int_{\mathbb R}p(y)p(y+r)\,dy,$$ whence $$P(d(X,Y)\le r)\sim2r\int_{\mathbb R}p(y)^2\,dy$$ as $r\downarrow0$.

Let $S:=\mathbb X$. Let $X$ and $Y$ be iid random elements of $S$ each with distribution $\mathbb Q$. It is more transparent to rephrase the question as follows:

Can one give a good lower bound on $P(d(X,Y)\le r)$?

Let $(A_i)$ be any countable measurable partition of $S$ with each $A_i$ of diameter $\le r$. Then clearly $$P(d(X,Y)\le r)\ge\sum_i P(X\in A_i,Y\in A_i)=\sum_i P(X\in A_i)^2,$$ so that $$P(d(X,Y)\le r)\ge L(r):=\sup_{(A_i)}\sum_i P(X\in A_i)^2,$$ where the $\sup$ is taken over all countable measurable partitions $(A_i)$ of $S$ with each $A_i$ of diameter $\le r$. The lower bound $L(r)$ on $P(d(X,Y)\le r)$ is clearly attained when $S$ is an at most countable metric space with $d(x,y)>r$ for all distinct $x,y$ in $S$. One can hardly have anything better in general.

If $S=\mathbb R$ with the usual distance for $d$ and if $X$ has (say) a bounded continuous pdf $p$, then $d(X,Y)$ has the pdf $q$ given by $$q(r)=2\,1(r>0)\int_{\mathbb R}p(y)p(y+r)\,dy,$$ whence $$P(d(X,Y)\le r)\sim2r\int_{\mathbb R}p(y)^2\,dy$$ as $r\downarrow0$.

Let $S:=\mathbb X$. Let $X$ and $Y$ be iid random elements of $S$ each with distribution $\mathbb Q$. It is more transparent to rephrase the question as follows:

Can one give a good lower bound on $P(d(X,Y)\le r)$?

Let $(A_i)$ be any countable measurable partition of $S$ with each $A_i$ of diameter $\le r$. Then clearly $$P(d(X,Y)\le r)\ge\sum_i P(X\in A_i,Y\in A_i)=\sum_i P(X\in A_i)^2,$$ so that $$P(d(X,Y)\le r)\ge L(r):=\sup_{(A_i)}\sum_i P(X\in A_i)^2,$$ where the $\sup$ is taken over all countable measurable partitions of $S$ with each $A_i$ of diameter $\le r$. The lower bound $L(r)$ on $P(d(X,Y)\le r)$ is clearly attained when $S$ is an at most countable metric space with $d(x,y)>r$ for all distinct $x,y$ in $S$. One can hardly have anything better in general.

If $S=\mathbb R^n$ with the Euclidean distance for $d$ and if $X$ has (say) a bounded continuous pdf $p$, then $d(X,Y)$ has the pdf $q$ given by $$q(r)=2\,1(r>0)\int_{\mathbb R^n}p(y)p(y+r)\,dy,$$ whence $$P(d(X,Y)\le r)\sim2r\int_{\mathbb R^n}p(y)^2\,dy$$ as $r\downarrow0$.

Let $S:=\mathbb X$. Let $X$ and $Y$ be iid random elements of $S$ each with distribution $\mathbb Q$. It is more transparent to rephrase the question as follows:

Can one give a good lower bound on $P(d(X,Y)\le r)$?

Let $(A_i)$ be any countable measurable partition of $S$ with each $A_i$ of diameter $\le r$. Then clearly $$P(d(X,Y)\le r)\ge\sum_i P(X\in A_i,Y\in A_i)=\sum_i P(X\in A_i)^2,$$ so that $$P(d(X,Y)\le r)\ge L(r):=\sup_{(A_i)}\sum_i P(X\in A_i)^2,$$ where the $\sup$ is taken over all countable measurable partitions of $S$ with each $A_i$ of diameter $\le r$. The lower bound $L(r)$ on $P(d(X,Y)\le r)$ is clearly attained when $S$ is an at most countable metric space with $d(x,y)>r$ for all distinct $x,y$ in $S$. One can hardly have anything better in general.

If $S=\mathbb R$ with the usual distance for $d$ and if $X$ has (say) a bounded continuous pdf $p$, then $d(X,Y)$ has the pdf $q$ given by $$q(r)=2\,1(r>0)\int_{\mathbb R}p(y)p(y+r)\,dy,$$ whence $$P(d(X,Y)\le r)\sim2r\int_{\mathbb R}p(y)^2\,dy$$ as $r\downarrow0$.