= Solution
Choose $n\geq1$ as in part (i). Since distances between iterates are nondecreasing, the <triangle inequality> gives
$$
d(\phi x,\phi y)\leq d(\phi^{(n)}x,\phi^{(n)}y)
\leq d(x,y)+d(x,\phi^{(n)}x)+d(y,\phi^{(n)}y)<d(x,y)+2\epsilon.
$$
Letting $\epsilon\downarrow0$ and combining with the given reverse bound proves that $\phi$ is an <isometry>:
$$
\boxed{d(\phi x,\phi y)=d(x,y)\quad(x,y\in X)}.
$$
For surjectivity, fix $x\in X$ and use part (i) with $y=x$ and $\epsilon=1/k$ to choose $n_k\geq1$ satisfying $d(x,\phi^{(n_k)}x)<1/k$. Put $z_k=\phi^{(n_k-1)}x$, interpreting $\phi^{(0)}$ as the identity. By <sequential compactness>, a <convergent subsequence> satisfies $z_{k_j}\to z\in X$. The assumed <continuity> yields
$$
\phi z=\lim_j\phi z_{k_j}=\lim_j\phi^{(n_{k_j})}x=x.
$$
Thus every $x$ has a preimage, and $\boxed{\phi\text{ is surjective}}$. If $X$ is empty, both conclusions hold vacuously. This proves the <distance-increasing self-map theorem for sequentially compact metric spaces>.
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