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 $x<1<y$, since $f$ is periodic. if $y=1+t$, we have $|f(x)-f(y)|=|f(x)-f(1+t)|=|f(x)-f(t)|\leq|f(x)-f(1)|+|f(1)-f(t)|=$ $=|f(x)-f(1)|+|f(0)-f(t)|<\varepsilon$ , because $|1-y|<\delta(\varepsilon/2)$ hence $|t|<\delta(\varepsilon/2)$ and $|1-x|<\delta(\varepsilon/2)$ too.
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?
It looks fine, but in case 2 you should have written $[0,1]$ instead of $(0,1)$.