For $p,q\in \mathbb R^3$, let $|p|$ and $p\times q$ denote respectively the Euclidean norm of $p$ and the cross product of $p$ and $q$. Define $d: \mathbb R^3\times \mathbb R^3 \to [0,\infty)$ by $$d(p,q)=|p|+|q| \text{ if } p\times q\ne 0 \\d(p,q)=|p-q| \text{ otherwise }$$
Prove that $d$ is a metric and show that the closed unit $d$-ball centered at $(0,0,0)$ isn't $d$-compact whereas the one centered at $(1,1,1)$ is.
To show this is a metric, the essential part is to prove the triangle inequality (other properties are obvious). I guess I need to consider 8 cases. Namely, given $p, q,r$, the cases are $p\times q=0\text{ or } \ne 0$ and similarly for $p\times r$ and $r\times q$. Is that correct? If it is, this part is clear (if we use $|p-q|\le |p|+|q|$).
For the balls part. The ball centered at the origin is the set of points with $d(0,x)\le 1$. Since $0\times x=0$, the points of the this ball satisfy $|0-x|\le 1$ or $x_1^2+x_2^2+x_3^2 \le 1$.
The points of the ball centered at $1:=(1,1,1)$ satisfy $d(1,x)\le 1$. So this ball is the set of points that satisfy $|x|+|1|\le 1$ (or equivalently $\sqrt{x_1^2+x_2^2+x_3^2}+\sqrt{3}\le 1$) together with the point $(0,0,0)$.
Are my descriptions correct so far? I don't know how to prove/disprove compactness. E.g. to disprove it, I need to find a collection of open sets w.r.t. this metric that cover the usual unit sphere and prove that no finite subcollection covers the sphere. But then I need to understand what open sets w.r.t. $d$ are to begin with. I know that in this case the base of topology is given by all open balls w.r.t. this metric, but I don't have any intuition to prove/disprove what is required.
In this metric, a closed ball of radius $r$ centered at $x$ is the set of all points $p$ with $\|p\|\le r-\|x\|$ along with all the points $\lambda x$ with $\lambda\in\mathbb R$ such that $|\lambda-1|\|x\|\le r$. (You missed the second part when describing the closed ball around $(1,1,1)$.) You can think of it as a ball centered at the origin with a line sticking out of it in the direction of $x$. For $r\le\|x\|$, it's just a line segment. The open balls are similar.
The line sticking out makes it possible to cover the unit closed ball around $(0,0,0)$ with open balls such that every point $x$ with $\|x\|=1$ is covered exactly once. This proves that this closed ball is not compact.
For the unit closed ball around $(1,1,1)$, notice that $S=\{\lambda(1,1,1):\lambda>0\}$ is an open set and it behaves very much like the positive real number line with the usual metric. The compactness of closed and bounded intervals will help you prove that this closed ball (which is a subset of $S$) is also compact. Can you take it from here?