Prove $f$ is uniformly continuous $\implies$ there exist $C, D$ such that $|f(x)| < C + D|x|$.
Proof below. Please verify or critique.
By definition of uniform continuity, there exists $\delta > 0$ such that $|x_a - x_b| \leq \delta \implies |f(x_a)- f(x_b)| < 1$. Choose $D > 1/\delta$ and $C > |f(0)| + D + 1$.
For any $x$, $|f(x)| - |f(0)| \leq |f(x) - f(0)| \leq \sum_{0 < j \leq |x/\delta|+1}|f(j\delta) - f((j-1)\delta)| \leq |x/\delta|+1$, so $|f(x)| \leq |x/\delta| + 1 + |f(0)| < C + D|x|$.