For an unbounded nonempty set of real numbers $D$, does there necessarily exist a continuous function $f$ that maps $D$ to the real numbers such that $f$ is not uniformly continuos?
No, there doesn't. Let $D = \mathbf N$, and let $f \colon \mathbf N \to \mathbf R$. We will show that $f$ is uniformly continuous: Let $\epsilon > 0$, let $\delta = \frac 12$. If $x,x' \in \mathbf N$ are given such that $|x-x'| < \delta = \frac 12$, we have $x=x'$, hence $f(x) = f(x')$, so $|f(x)-f(x')| < \epsilon$.