= Solution
Here <equivalent metrics> means that the two <metrics> induce the same <topology>, and <Lipschitz equivalent norms> means that there are constants $c,C>0$ such that $cp(x)\le q(x)\le Cp(x)$ for every $x$.
Suppose first that the inequalities hold. Since they also hold for $x-y$, the identity maps between the two <metric spaces> are <Lipschitz continuous>. In particular, each is continuous, so the <topologies> agree.
Conversely, equality of the <topologies> makes the identity from the $p$-<norm> to the $q$-<norm> continuous at zero. There is therefore $\delta>0$ such that $p(x)<\delta$ implies $q(x)<1$. For nonzero $x$, apply this to $\delta x/(2p(x))$ and use the absolute homogeneity of both <norms>:
$$
q(x)<\frac{2}{\delta}p(x).
$$
The reverse identity is also continuous, so there is $\varepsilon>0$ with $q(x)<\varepsilon\Rightarrow p(x)<1$. Rescaling again gives $p(x)<2q(x)/\varepsilon$. Thus
$$
\boxed{\frac{\varepsilon}{2}p(x)\le q(x)\le\frac{2}{\delta}p(x).}
$$
The inequalities also hold at zero. This is <topology determines norm equivalence>; it works in arbitrary dimension because the rescaling argument needs neither a <basis> nor compactness of a unit sphere.
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