I began this way
Assume we have a periodic continuous function $f$. That is, there exists a $\delta$ such that if $0<|x-x_0|<\delta$, then $|f(x)-f(x_0)|<\varepsilon$ and $f(x+h)=f(x)$ for all $x$ and some $h\neq 0$. We wish to show that $f$ is uniformly continuous. That is, there exists a $\delta$ such that $0<|x-y|<\delta$ implies that $|f(x)-f(x+y)|<\varepsilon$ where $\varepsilon>0$.
I know this seems like a duplicate but I have looked at the solutions and hints already posted and it is not. I need to do this without HB or anything past it. Thanks for your help!