We know the definition of Uniform-continuity is
If $A \subset \mathbb{R}$ and $f:A \rightarrow \mathbb{R}$ with $$\forall \epsilon>0\ (\exists \delta>0\ (|x-y|<\delta \implies |f(x)-f(y)|<\epsilon) )$$ then f is uniformly continuous on A.
If we consider the statement: Given $f:A \rightarrow \mathbb{R}$ such that $$\forall \delta>0\ (\exists \epsilon>0\ (|x-y|<\delta \implies |f(x)-f(y)|<\epsilon) )$$
Does this statement also hold for uniformly continuous functions defined on $A$?
The latter implication follows purely logically from the first.
That implies, yes, the property is valid for all uniform continuous functions.
So for non uniform-continuous functions: A counter example is $\mathbb R - \{0\}\ni x \mapsto 1/x$ . Assume there is such a $\delta$, then set $x=-\delta/2$, $y=\delta/2$ Then $d(x,y)=\delta$, but $d(f(x),f(y))$ is not bounded for $\delta$ arbitrarily small