Proof: By uniform continuity we can choose $\delta>0$ such that $|x-y| \leq \delta$ implies $ |f(x)-f(y)|<1$ .

Let $x \geq a$ , choose $n$ to be the smallest positive integer such that $x \leq (n+1)\delta$ . Then $$ f(x) -f(a) = f(x) - f(a+n\delta) + \sum_{i=1}^n f(a+i\delta)-f(a+(i-1)\delta) $$ so that we have \begin{eqnarray} |f(x)| &\leq& |f(x)-f(a+n\delta)| + \sum_{i=1}^n |f(a+i\delta)-f(a+(i-1)\delta)| + |f(a)| \\ &\leq& n+ 1+ |f(a)|. \end{eqnarray}Therefore, \begin{eqnarray} \frac{|f(x)|}{x} &\leq& \frac{|f(a)| + n+1}{n\delta}\\ &\leq& \frac{|f(a)|}{n\delta} + \frac{n+1}{n\delta}. \end{eqnarray}As $n\to \infty$ , the rhs converges to $\frac{1}{\delta}$ . Hence, the sequence defined by $b_n = \frac{|f(a)|}{n\delta} + \frac{n+1}{n\delta}$ is bounded by some number $B$ as desired.

Note we can extend this result to $f:[0,\infty)\to \mathbb{R}$ if $f$ is differentiable at 0.