There are infinitely many vectors such that $\|\mathbf u-(1,1,1)\|\le 3$ and $\|\mathbf u+(1,1,1)\|\le 3$ I want to prove that there are infinitely many solutions in $3$-space for $\|\mathbf u-(1,1,1)\|\le 3$ and $\|\mathbf u+(1,1,1)\|\le 3$ (where bold refers to vectors). My proof: 

What is wrong with my current proof? If you can prove this question please  post your proof. 
 A: You want to prove the intersection of the closed balls of radius $3$ around $(1,1,1)$ and $(-1,-1-1)$ has infinite cardininality.
It suffices to prove the intersection of the open balls of radius $3$ around $(1,1,1)$ and $(-1,-1-1)$ is non-empty (because non-empty open sets have infinite cardinality). To see this notice $(0,0,0)$ is in this intersection.
A: Consider the plane $P=\{(x,y,z): x+y+z=0\}$ which is orthogonal to the line thru  $(1,1,1)$ and $(-1,-1,-1).$ For $q\in P$ and $q\ne (0,0,0)$, and  if $r=(1,1,1)$ or if $r=(-1,-1,-1)$, then the points $q,r,$ and $\bar 0=(0,0,0)$ are the vertices of a right triangle with the angle at $\bar 0$  being $90^o.$ Consider any circle $S$ on $P,$ centered at $\bar 0,$ with radius $c>0.$ The distance from any point $q$ on $S,$ to $(1,1,1),$ or to $(-1,-1,-1),$ is $\sqrt {c^2+3}.$ This is less than $3$ whenever $c<\sqrt 6.$  Personally I like dREaM's solution more,but he beat me to it.
A: Geometrically, you are looking at the intersection of two balls.
If you know some basic point-set topology, then you know this is an open set and hence contains a ball, which has infinitely many points.
Maybe you want a more concrete argument. Notice that the origin is the midpoint of $(1, 1, 1)$ and $(-1, -1, -1)$. Each center is a distance $\sqrt 3$ to the origin. If $(x,y,z)$ is a point in the unit ball (i.e. $\|(x,y,z)\|< 1$), then the distance from $(1,1,1)$ to $(x,y,z)$ is less than $1+\|(x,y,z)\|$ (use the triangle inequality). So the unit ball is contained in the intersection.
Now you only need to show that the unit ball contains infinitely points. For this, notice that there are infinitely many numbers $\lambda \in (-1,1)$ and $(\lambda, 0, 0)$ is in the unit ball. You've found infinitely many points on just one line (the $x$-axis), there are plenty more in the unit ball!
A: You proved the following statement:

If $\|u-(1,1,1)\| \le 3$ and $\|u+(1,1,1)\| \le 3$, then $\|u\| \le \sqrt{6}$. 

The converse of that statement: 

If $\|u\| \le \sqrt{6}$ then  $\|u-(1,1,1)\| \le 3$ and $\|u+(1,1,1)\| \le 3$

is not true. 
If you try to reverse your steps, you will see that $\|u\| \le \sqrt{6}$ implies $\|u-(1,1,1)\|^2+\|u+(1,1,1)\|^2 \le 18$. But this does not imply that $\|u-(1,1,1)\|^2 \le 9$ and $\|u+(1,1,1)\|^2 \le 9$. 
As you pointed out by testing $u = (1,1,1)$, it is possible to have $u$ satisfy "$\|u-(1,1,1)\|^2+\|u+(1,1,1)\|^2 \le 18$" and not satisfy "$\|u-(1,1,1)\|^2 \le 9$ and $\|u+(1,1,1)\|^2 \le 9$". 

To prove that infinitely many $u$ satisfy $\|u-(1,1,1)\| \le 3$ and $\|u+(1,1,1)\| \le 3$, first, show that there are infinitely many vectors $u$ such that $\|u\| \le 3-\sqrt{3}$. Then, by the triangle inequality, all of those vectors satisfy $\|u-(1,1,1)\| \le \|u\|+\|-(1,1,1)\| \le (3-\sqrt{3})+\sqrt{3} = 3$, and $\|u+(1,1,1)\| \le \|u\|+\|(1,1,1)\| \le (3-\sqrt{3})+\sqrt{3} = 3$. Therefore, there are infinitely many vectors $u$ which satisfy $\|u-(1,1,1)\| \le 3$ and $\|u+(1,1,1)\| \le 3$.
