Let $f$ be continuous function on $[0,\infty)$. If $f$ is uniform continuous on $[k,\infty)$ for some $k$ then $f$ is uniform continuous on $[0,\infty)$
I start proof as:
Now f is uniform continuous on $[K,\infty)$ and also on $[0,k+1]$.
So we get the following for $ \epsilon>0$:
When I tried it myself I took $\delta= \min\{\delta_1, \delta_2\}$ why do we take 1 in this set not able to understand? Please help me.
You need that $x,y$ are both in $[0,k+\color{red}1]$ or both in $[k,\infty)$. You can conclude this only if $|x-y|\le 1$.
Alternatively, you might have worked with $[0,k+\delta_1]$ instead of $[0,k+1]$.