real analysis - 1-periodic and continuous is uniformly continuous

Let $f$ be a $1$-periodic function that is continuous on $\mathbb{R}$. Then $f$ is uniformly continuous on $\mathbb{R}$.

Indeed, since $f$ is continuous on $\mathbb{R}$, it is continuous in $[0,1]$ which is compact; therefore $f$ is uniformly continuous there, that is, for $\varepsilon>0$ there exists $\delta(\varepsilon)>0$ such that for all $x,y \in [0,1]$ $|x-y|<\delta\implies|f(x)-f(y)|<\varepsilon$. Now, let $x,y$ be two points of $\mathbb{R}$ such that $|x-y|<\delta(\varepsilon/2)$. We distinguish two cases:

Case 1: there exists an integer $n$ between $x,y$. Then without loss of generality, we may assume that $x<1

Case 2: there is no integer between $x,y$. Then without loss of generality (periodicity) we may assume that $x,y\in (0,1)$ and we have nothing to prove.

Any objections?