Baby Rudin Exercise 1.16 The following is exercise problem 16 from chapter 1 of Principles of Mathematical Analysis by Rudin.

Suppose $k \ge 3, \mathbb{x,y} \in R^k, \mid\mathbb{x-y}\mid=d>0,$ and $r>0.$ Prove that if $2r>d,$ there are infinitely many $\mathbb{z}\in R^k$ such that $$\mid\mathbb{z}-\mathbb{x}\mid =\mid\mathbb{z}-\mathbb{y}\mid = r. $$

My attempt :
Suppose I sketch a cross section for $k=3$ case, then I would have the following figure :

Here $AB=d$ and $AM=d/2.$ Also $AE=BE=r$(where $r>d/2$) so that we have spheres of radius $r$ centred at $A$ and $B.$
From the figure we observe that $\mathbb{z}=\mathbb{p}+\mathbb{w}$ is one of the solution satisfying the conditions where $\mathbb{w}$ should be such that 
$$\mathbb{w}\cdot (\mathbb{x}-\mathbb{y})=0$$ $$|\mathbb{w}|^2= r^2 - (d/2)^2$$
Thus if we define $\mathbb{z}$ as above then the set of all such $\mathbb{z}$ will form a solution set. My question is how do I proceed from here to show that this solution set is infinite?

This is a problem of dimensions. Note that it was assumed $k\geq3$. Unfortunately you have introduced new notations. From your picture it becomes apparent that $M$ is the central point of the problem. We therefore let $M$ be the origin $0$ of our system. This means that $y=-x\ne0$. The line through $x$ and $y$ has an orthogonal hyperplane $H\subset{\mathbb R}^k$  through $0$, whose dimension is $k-1\geq2$. When $r>{|x-y|\over2}=|x|$ there is a small sphere of dimension $k-2\geq1$ in $H$ of radius $\rho=\sqrt{r^2-|x|^2}>0$. This sphere has infinitely many points, and all of these points have distance $r>0$ from $x$ and $y=-x$.
