Let $\langle X,\rho \rangle$ be a metric space and $f:\emptyset\to X$ a function. Since $\emptyset$ is compact, I know that $f$ is uniformly continuous. But can it be proven by vacuous truth? It's the same argument as for continuity, or something change because the alteration of the quantification?
The alteration of quantification doesn't affect the vacuous truth here. To see this, write down the definition of uniform continuity:
$$ \forall \epsilon, \exists \delta, \forall x, y \in \emptyset : |x - y| < \delta \implies |f(x) - f(y)| < \epsilon $$
For a fixed $\epsilon > 0$, any $\delta > 0$ would work. The implication $|x - y| < \delta \implies |f(x) - f(y)| < \epsilon$ is vacuously true because there are no $x, y$ to satisfy $|x - y| < \delta$ in $\emptyset$ no matter what $\delta$ is chosen.