Assume $\varrho(x,y)<\delta \implies \varrho[(h(x),h(y)]<\varepsilon$. Then for all points $h(y)\in h[B_\delta (x)]$, i.e. in the image of $B_\delta(x)$, which consists of all point $y$ for which it is true that $\varrho(y,x)<\delta$, it is true by assumption that $\varrho[h(x),h(y)]<\varepsilon$, which means that $h(y) \in B_\varepsilon(h(x))$, so $h[B_\delta (x)] \subseteq B_\varepsilon(h(x))$.
Assume $h[B_\delta (x)] \subseteq B_\varepsilon(h(x))$. Then $h(y) \in h[B_\delta (x)]$ implies $h(y) \in B_\varepsilon(h(x))$. The former is true iff the pre-image of $h(y)$ is in the pre-image of $h[B_\delta (x)]$, i.e. $y \in B_\delta (x)$, meaning $\varrho(x,y)<\delta$. And from the latter we have that $\varrho[h(x),h(y)] <\varepsilon$.
Is it correct enough?
It's close enough. If you think about it, it's an almost trivial reformulation.
The write-up could be a little bit better. My proposal:
Suppose the implication holds.
Let $y \in h[B_\delta(x)]$. This means that $y=h(x')$ with $x' \in B_\delta(x)$, which means $d(x,x') < \delta$. The assumed implication gives us that we thus have $\rho(h(x), h(x')) < \varepsilon$. The latter implies that $y=h(x') \in B_\rho(h(x))$. This shows the inclusion.
The reverse is almost the same.