I proved if $f : \mathbb R\to \mathbb R$ is periodic and $C^1$ class, then $f$ is Lipschitz continuous.
I wonder if my proof is correct and if there is an easier proof. My proof is a little complicated.
My Proof
Let the period $T$.
Since $f$ is periodic and continuous, $f$ is bounded so there exists $M>0$ s.t. $|f(x)|\leqq M$ for all $x \in \mathbb R.$
Now, define $g:=f|_{[-T,T]}$.
Since $f\in C^1$, $g'$ is continuous on the compact set $[-T, T]$ thus there exists $K>0$ s.t. $|g'(x)|\leqq K$ for all $x \in [-T,T]$
Let $L:=\max \{ \frac{2M}{T}, K \}.$ I'll prove $|f(x)-f(y)|\leqq L|x-y|$ for all $x,y \in \mathbb R.$
Let $x,y \in \mathbb R.$ I can assume $x>y.$
If $x-y\geqq T$, then $|f(x)-f(y)|\leqq |f(x)|+|f(y)| \leqq 2M = \frac{2M}{T} T \leqq \frac{2M}{T} (x-y)\leqq L|x-y|.$
Consider the case of $0<x-y<T.$
There exist $n,m \in \mathbb Z$ s.t. $(2n-1)T \leqq x < (2n+1)T,\ (2m-1)T \leqq y < (2m+1)T$.
Then, $(2n-2m-2)T<x-y< (2n-2m+2)T$.
Since $0<x-y<T,$ $0<(2n-2m+2)T$ and $(2n-2m-2)T<T$ have to hold, thus I get $n-m=0,1.$
If $n-m=0,$
\begin{align} |f(x)-f(y)| &=|f(x-2nT)-f(y-2nT)| \ (\mathrm{the \ periodicity})\\ &=|g(x-2nT)-g(y-2nT)| \ (-T\leqq y-2nT< x-2nT<T) \\ &=|g'(c)||(x-2nT)-(y-2nT)|\ (\mathrm{MVT}, \exists c\in (y-2nT, x-2nT))\\ &\leqq K|x-y| \\ &\leqq L|x-y|. \end{align}
Consider the case of $n-m=1$.
Then, $(2m+1)T\leqq x<(2m+3)T,\ (2m-1)T\leqq y< (2m+1)T$.
And $x=x-y+y<T+y<(2m+2)T,\ y=y-x+x >-T+x\geqq 2mT.$
Thus $2mT<y<(2m+1)T \leqq x<(2m+2)T$ and therefore $x-(2m+1)T, y-(2m+1)T \in [-T,T]$.
Hence \begin{align} |f(x)-f(y)| &=|f(x-(2m+1)T)-f(y-(2m+1)T )|\\ &=|g(x-(2m+1)T)-g(y-(2m+1)T )| \\ &=_{\mathrm{MVT}}|g'(d)||(x-(2m+1)T)-(y-(2m+1)T )| \\ &\leqq K |x-y|\\ &\leqq L|x-y|.\end{align}
Therefore $|f(x)-f(y)|\leqq L|x-y|$ for all $x,y \in \mathbb R.$
Is this proof correct ? And is there an easier way to prove ?
It is not difficult to show that if $f$ is differentiable on $\mathbb{R}$, then $$\sup\{\frac{|f(x) - f(y)|}{|x - y|} : x, y \in \mathbb{R}\} = \sup\{|f'(x)| : x \in \mathbb{R}\}.$$ Hence showing $f$ is Lipschitz is equivalent to showing $f'$ is bounded. If $f$ is periodic and $C^1$, then $f'$ is periodic and continuous. Any periodic continuous function is bounded by compactness, so you have your result.