If $f$ is continuous on $[0, \infty)$ and uniformly continuous on $[b, \infty)$ for some $b > 0$ then $f$ is unif. continuous on $[0, \infty). Prove that if $ f $ is continuous on $ \left[ 0, \infty\right)$ and uniformly  continuous on $[b, \infty)$ for some $b > 0$ then $f$ is unif. continuous on $[0, \infty)$.
So far I have: Let $A = [b, \infty)$ then $A^c = [0,b)$, my goal is to end up with a compact set from $[0,b]$ so I can use the theorem that "A function is continuous on a compact set $K$ is uniformly continuous on $K$" and then take the fact that of the preservation of connected sets. Am I going about this in the right way?

A continuous function on a compact set is uniformly continuous.  Since $f$ is uniformly continuous on $[0, b]$ and uniformly continuous on $[b, \infty)$, it is uniformly continuous on $[0, \infty)$.  
To see that it is uniformly continuous, let $\epsilon > 0$.   We can find $\delta_1, \delta_2 > 0$ such that $|x - y| < \delta_1, \delta_2$ implies $|f(x) - f(y)| < \epsilon$ for $x, y \in [0, b]$ and $x, y \in [b, \infty)$ respectively.  Set $\delta = \min\{\delta_1, \delta_2\}$.  Then for any $x \in [0, b]$ and $y \in [b, \infty)$, if $|x - y| < \delta$ then
$$|f(x) - f(y)| < |f(x) - f(b)| + |f(b) - f(y)| < 2\epsilon$$
Since $\epsilon > 0$ was arbitrary, this completes the proof.
