Is this a polyhedron? 
Is $S$ a polyhedron?
$$S=\{x\in\mathbb{R}^n|\|x-x_0\|\le\|x-x_1\|\}$$ where $x_0, x_1$ are
  given. $S$ is the set of points that are closer to $x_0$ than to
  $x_1$.

I was thinking the answer is yes, since $S$ is a halfspace, but how do I show that $S$ is a halfspace?
 A: To get to know this set, we express the norms using inner products.
$$
\|x-x_0\|\le\|x-x_1\| \Leftrightarrow
\|x-x_0\|^2\le\|x-x_1\|^2 \Leftrightarrow
\langle x-x_0, x-x_0 \rangle\le \langle x-x_1,x-x_1\rangle.
$$
We have
$$
\langle x-x_0, x-x_0\rangle = \langle x,x\rangle - 2\langle x,x_0\rangle + \langle x_0,x_0\rangle,
$$
thus our inequality becomes
$$
-2\langle x,x_0\rangle+\langle x_0, x_0\rangle \le -2\langle x,x_1\rangle+\langle x_1,x_1\rangle
$$
which is equivalent to
$$
0 \le \langle x,2(x_0-x_1)\rangle+\|x_1\|^2-\|x_0\|^2,
$$
or even
$$
0 \le \langle x,x_0-x_1\rangle+\frac{1}{2}\left(\|x_1\|^2-\|x_0\|^2\right).
$$
For $=$ instead of $\le$ this describes a hyperplane with normal vector $x_0-x_1$ going through the midpoint $\frac{1}{2}(x_0+x_1)$. Therefore our inequality describes a halfspace with the described hyperplane as its boundary.
A: Notice $$\|x-x_0\|\le\|x-x_1\| \iff \|x-x_1\|^2 - \|x-x_0\|^2 \ge 0$$ and the quadratic term in the LHS of last expression cancel each other. We find
$$S = \big\{\; x \in \mathbb{R}^n : 2 (x_0 - x_1) \cdot x + \|x_1\|^2 - \|x_0\|^2 \ge 0\;\big\}$$
This is the equation of a half-space.
