For a metric space (iii) is equivalent to being compact and implies (i) (a compact metric space is bounded) and (ii) (the continuous image of a compact set is compact). (ii) obviously implies (i) (fix a point, the distance from the point is continuous). Now, (ii) also implies (iii): if an infinite subspace has no limit points then it is closed and discrete; a surjective function from it on the set of positive integers is continuous and can be extended to a continuous unbounded function (the space is metrizable, therefore, normal). We need only to show that (i) implies (ii) or (iii). We show (i) implies (ii). The hint was ambiguous for me so I figured out an alternative proof.Remember from Section 20 that the topology generated by a metric is the coarsest topology such that the metric as a function is continuous. This implies also that if a metric $d'$ is continuous relative to $(X,d)$ then $(X,d')$ is coarser than $(X,d)$ . We prove the following lemma.Lemma. Suppose that $(X,\mathcal{T})$ is metrizable and every metric that generates the topology is bounded. Then every metric $d'$ continuous relative to $\mathcal{T}$ is bounded as well.Proof. Suppose $(X,d)=(X,\mathcal{T})$ . $d$ is bounded. Let $m=d+d'$ . $m$ is a metric (easily verifiable) and continuous relative to $(X,d)$ , therefore, $(X,m)$ is coarser than $(X,d)$ . We show that, in fact, $(X,m)=(X,d)$ . For every $y\in B_d(x,r)$ , suppose $z\in B_m(y,r-d(y,x))$ , then $d(z,x)\le d(x,y)+d(y,z)\le d(x,y)+m(y,z)<r$ , therefore, for every point in $B_d(x,r)$ there is a neighborhood in $(X,m)$ contained in the ball, and $(X,m)$ is finer than $(X,d)$ . This implies that $m$ is bounded, and so is $d'$ .Now, take any continuous $\phi$ . And suppose $d$ is a metric on $X$ . Then $d(x,y)+|\phi(x)-\phi(y)|$ is a metric that is continuous relative to the topology on $X$ . Therefore, it is bounded, and so is $\phi$ .