= Solution
By the permitted <sequential compactness> of the <product metric> space $X\times X$, the sequence of pairs $(\phi^{(k)}x,\phi^{(k)}y)$ has a <convergent subsequence>, and hence a <Cauchy subsequence>. Choose two of its indices $r<s$ sufficiently late that
$$
d(\phi^{(r)}x,\phi^{(s)}x)+d(\phi^{(r)}y,\phi^{(s)}y)<\epsilon.
$$
The distance-increasing assumption iterates: $d(\phi^{(r)}u,\phi^{(r)}v)\geq d(u,v)$. With $n=s-r\geq1$, use $u=x,v=\phi^{(n)}x$, and then $u=y,v=\phi^{(n)}y$, to obtain
$$
d(x,\phi^{(n)}x)+d(y,\phi^{(n)}y)\leq d(\phi^{(r)}x,\phi^{(s)}x)+d(\phi^{(r)}y,\phi^{(s)}y)<\epsilon.
$$
Both summands are nonnegative, so
$$
\boxed{d(x,\phi^{(n)}x)<\epsilon,\qquad d(y,\phi^{(n)}y)<\epsilon}.
$$
The use of paired iterates ensures that the same positive return time works for both points.
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