If $B_{r}(x)\subseteq B_{s}(y)$ then what can be said about $\vert \vert x-y \vert \vert$ Let $B_{r}(x)\subseteq B_{s}(y)$ where $r,s>0$ and $x$ and $y$ are arbitrary points. My guess is that (following a number a drawings) $\vert \vert x-y \vert \vert<s-r$ but I am struggling to prove it mathematically. 
My idea: I think the best way to go about it would be via contradiction, so assume $\vert \vert x-y \vert \vert\geq s-r$ that would then mean that $x-y \notin B_{s-r}(0)$ and note that for $z:=z_{1}-z_{2}\in B_{s-r}(0)$ it is necessary that $z_{1} \in B_{s}(0)$ and $z_{2} \in B_{r}(0)$
Therefore, by consequence $x \notin B_{s}(0)$ or $y\notin B_{r}(0)$
I feel like I am on the wrong track. Any ideas
 A: Your conjecture is not correct, take $x=y$ and $r=s$ then your conjecture would say that $0<0$, which is obviously false. However, we can prove that $\| x-y \|\le s-r$.
The assumption $B_{r}(x)\subseteq B_{s}(y)$ means that for any given $x$ such that $\|x-x'\|<r$, we have $\|y-x'\|<s$. We will show by contradiction that if $\| x-y \| = d> s-r$, then we can exhibit $x'$ that violates the above.
Let $z:=x-y$ and $\varepsilon = d-(s-r)$. Take $x'=y+\left(1+\frac{r-\varepsilon/2}{\|z\|}\right)z$. We have
$$
x'-x =\frac{r-\varepsilon/2}{\|z\|} z
$$
so $\|x-x'\|=r-\varepsilon/2 < r$. However, 
$$\begin{align}
x'-y &= (\|z\| +r-\varepsilon/2)  \frac{z}{\|z\|} \\
&= (d+r-\varepsilon/2)\frac{z}{\|z\|} \\
&= (s+\varepsilon/2)\frac{z}{\|z\|},
\end{align}$$
hence $\|y-x'\|=s+\varepsilon/2 > s$, a contradiction.
A: If $x=y$ there is nothing to prove. Otherwise, consider the line $l=\{t(x-y): t\ge 0\}$. Let $z$ be the point in $l\cap B_r(x)$ with maximum norm (I am taking the closed ball, so there is such a point by compactness and continuity of the norm). Then, since $y,x,z$ are on $l$,
$$
\|x-z\|+\|y-x\|= \|y-z\|\le s,
$$
because $z\in B_r(x)\subset B_s(y)$. Since $\|x-z\|=r$ you are done. (The right thing is $\le$ and not $<$)
