Proof verification: if $f$ is UC on $[k,\infty)$ for some $k$, then f is UC on $[0,\infty)$ This is a question from Elementary Analysis by Ross.

Let $f$ be a continuous function on $[0, \infty)$. Prove that if $f$
  is uniformly continuous on $[k,\infty)$ for some $k$, then $f$ is
  uniformly continuous on $[0,\infty)$.

Here's my attempt of a proof:
If $k\le 0$ then we are done. So, we assume that $f$ is uniformly continuous on $[k, \infty)$ for some $k>0$. If it were the case that $f$ was not uniformly continuous on $[0, \infty)$ then $f$ would not be uniformly continuous on $[k, \infty)$. Hence, it must be that $f$ is uniformly continuous on $[0, \infty)$.
Is this proof correct?
 A: Your argument is wrong because you argue that being uniformly continuous on a smaller set implies uniform continuity on a bigger set, which is in general false. See my comment above for a counter example. Here's a correct proof.
Let $f$ be continuous on $[0,\infty)$ where $f$ is uniformly continous on $[k,\infty]$. for some $k \in \mathbb{R}$. Then let $\epsilon > 0$, then there exists a $\delta > 0$ such that if $x,y \geq k$ and $|x-y| < \delta$, then $|f(x)-f(y)| < \epsilon/4$.
Now as $f$ is continuous on $[0,k]$, then we see by Weiestrass's theorem that $f$ is uniformly continous on $[0,k]$ as $[0,k]$ is compact by Heine-Borel. Hence, there exists an $\delta_{1} > 0$ such that if $x,y \in [0,k]$, then $|f(x)-f(y)| < \epsilon/4$
Now let $\delta_{2} = \min(\delta, \delta_{1})$
Then for any $x,y \in [0,\infty)$, where $|x-y| < \delta_{2}$ (if both $x,y \in [0,k$] or $[k,\infty]$, then by above we are done, so we only need to consider the intersection.
WLOG assume $x \in [0,k]$, $y \in [k,\infty)$. 
$|f(x)-f(y)| \leq |f(x)-f(\zeta)| + |f(\zeta)-f(k)| + |f(k) - f(\chi)| + |f(\chi) - f(y)|$ 
where $\zeta \in [0,k]$ such that $|\zeta-x| < \delta_{2}, |x-k| < \delta_{2}$ and similarly for $\chi$. Note we can choose such a $\chi, \zeta$ because $x \in [0,k]$, $y \in [k,\infty)$ and $|x-y| < \delta_{2}$
Collecting these we have $|f(x)-f(y)| < 4*\epsilon$/4 = $\epsilon$. 
