Prove $s\leq r-d(Q,P)\implies B(Q,s)\subseteq B(P,r)$
where $B(P,r)$ is the set of points inside a circle with radius $r$ centered at $P$ and $d(P,Q)$ is the distance between $P$ and $Q$. So
$$B(P,r)=\{x\mid d(P,x)\leq r\}$$
Assuming $s\leq r-d(Q,P)$:
$$d(Q,P)\leq r-s$$
I started by proving
$$B(Q,s)\subseteq B(Q,r)$$ $$B(P,s)\subseteq B(P,r)$$
and
$$Q\in B(P,r-s)\subseteq B(P,r)$$ $$P\in B(Q,r-s)\subseteq B(Q,r)$$
but I don't know if that's the right path.
You can do it in a simpler way, which is the standard way to prove inclusion. I'll show you here:
Suppose $s\le r-d(Q,P)$. Then let $y\in B(Q,s)$.
We are gonna show that $y\in B(P,r)$. To do this, let's see if $d(y,P)\le r$.
$$y\in B(Q,s)\implies d(y,Q)\le s\le r-d(Q,P)\implies d(y,Q)+d(Q,P)\le r$$ But from the triangular inequality property of $d$ we know that $d(y,P)\le d(y,Q)+d(Q,P)$, and using the inequality showed before we get $d(y,P)\le d(y,Q)+d(Q,P)\le r\implies y\in B(P,r)$