= Solution
Use <closed balls> with centres belonging to the space being covered. The <covering number> is
$$
\boxed{N(\varepsilon;R,\rho)=\min\left\{|A|:A\subseteq R,\ R\subseteq\bigcup_{a\in A}\overline B_\rho(a,\varepsilon),\ |A|<\infty\right\},}
$$
with value $\infty$ if there is no finite <metric net>, and value zero for the empty space.
With this intrinsic-centre convention, \b[the proposed monotonicity under subsets is false]. Give $R=\{-1,0,1\}$ the usual distance, take $S=\{-1,1\}$, and set $\varepsilon=3/2$. A ball centred at $0$ covers $R$, so $N(3/2;R,\rho)=1$. The two points of $S$ are distance $2$ apart, so a ball centred in $S$ contains only its centre, and $N(3/2;S,\rho)=2$. These values are also valid with <open balls>.
The obstruction is that the useful centre $0$ disappears on passing to the subspace. If centres were allowed anywhere in a fixed ambient space, monotonicity would hold. With intrinsic centres, one instead has
$$
\boxed{N(2\varepsilon;S,\rho)\leq N(\varepsilon;R,\rho).}
$$
Indeed, discard covering balls missing $S$ and choose one point of $S$ from each remaining ball. The <triangle inequality> shows that every point of its intersection with $S$ is within $2\varepsilon$ of the chosen point. This proves the replacement bound whenever the right side is finite; otherwise it is automatic.
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