$f$ is uniformly continuous $\implies |f(x)| < C + D|x|$ 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 $|f(x)| \leq |x/\delta| + 1 + |f(0)| < C + D|x|$.
 A: It seems to be that you are assuming $x$ to be positive and you are trying to partition $[0, x]$ and use triangle inequalities. Be careful that $x$ can be negative and you are partitioning the interval correctly.
Let's define partition on $[0,|x|]$ such that each of the segments are at most length $\delta$.
Let $m = \left\lfloor \frac{|x|}{\delta} \right\rfloor$, $$m \le \frac{|x|}{\delta} < m+1$$
$$m \delta \le |x| < (m+1)\delta.$$
Let's divide the interval to $m+1$ segments with
$$x_i = \begin{cases} i \delta, & i \le m \\ |x|, & i = m+1\end{cases}$$
Let $s= sign(x)$,
\begin{align}|f(x)|-|f(0)| &\le |f(x)-f(0)| \\&\le \sum_{i=0}^{m}|f(sx_{i+1})-f(sx_i)|\\&\le (m+1)\\
&\le \left(\frac{|x|}{\delta}+1 \right)  \end{align}
Also, I think you can just choose $C>|f(0)|+1$ though you can pick a larger number.
A: I think what you do is correct. A minor correction would be instead of $j \in \{i : 0 \le i \le \lfloor|x|/\delta\rfloor+1\}$, you could sum over $j \in \{i : 0\le  i \cdot sgn(x) \le \lfloor|x|/\delta\rfloor+1\}$. This way it works for negative $x$ values as well.
