My analysis text defines uniform continuity as
Definition. A function $f: A \to \mathbb{R}$ is uniformly continuous on a set $A \subseteq \mathbb{R}$ if for any $\epsilon > 0$, there exists a $\delta > 0$ such that for all $x,y \in A$, $|x-y| < \delta$ implies that $|f(x) - f(y)| < \epsilon$.
Since $\emptyset \subseteq \mathbb{R}$, does this satisfy the definition of uniform continuity vacuously? I think that both conditions are satisfied since there is no element of the empty set to test this criterion on and $f(\emptyset) = \emptyset$
$\DeclareMathOperator{\im}{Im}$ Yes, even a real function $f: \mathbb{R} \to \mathbb{R}$ that is discontinuous at every point is uniformly continuous over the empty set by a vacuous truth.
Generally, an arbitrary real valued function $f$ is uniformly continuous over any finite set $X \subseteq \mathbb{R}$ over which $f$ is defined because the image of the function over $X$ is finite and therefore bounded, so there is always some value of $\delta > 0$ such that the only value of $y$ that satisfies $|x - y| < \delta$ is $x$, resulting in $|f(x) - f(y)| = 0 < \epsilon$.