Suppose that $f$ has period one. Since $f:[0,2]\to\Bbb R$ is continuous, it is bounded, so $f$ is bounded all over $\Bbb R$ (why?). Also, $f:[0,2]\to\Bbb R$ is uniformly continuous, being continuous on a compact set. Thus, given $\varepsilon >0$ there exists $\delta>0$ such that, whenever $|x-y|<\delta,x,y\in[0,2]$, then $|f(x)-f(y)|<\varepsilon$. Pick arbitrary $x,y$, and assume $x<y$ with $|x-y|<\delta$. We may take $\delta <1$. I claim there is an integer $n$ such that $x-n,y-n\in [0,2]$. Then $|x-y|=|x-n-(y-n)|$ and $f(x-n)=f(x)$, $f(y-n)=f(y)$. Can you continue now?
Drawing a picture would prove useful. Essentially, you're translating the problem to $[0,2]$ where we already solved the issue.
Suppose that $f\colon \def\R{\mathbf R}\R \to \R$ is continuous, bounded and monotone. Then $f$ is uniformly continuous.
Suppose wlog that $f$ is monotonically increasing. Let $s := \sup f$ and $i := \inf f$. Let $\epsilon > 0$, choose $L > 0$ such that
$$ f(x) > s-\epsilon, \qquad x \ge L $$
and
$$ f(x) < i +\epsilon, \qquad x \le -L$$
Note that $f$ is uniformly continuous on $[-L-1, L+1]$ choose $\delta_0$ such that
$$ |x-y| <\delta_0, |x|,|y| < L+1 \implies |f(x)- f(y)| < \epsilon $$
Now let $\delta := \min(1, \delta_0)\le \delta_0$. Let $x,y\in \R$ be arbitrary. If $x \ge L+1$, then $y \ge L$, hence both $x,y\ge L$, therefore $f(x),f(y) \in [s-\epsilon,s]$, that is
$$ |f(x)-f(y)| \le \epsilon $$
Doing this same line of thought again we obtain
$$ |f(x) - f(y)| \le \epsilon, \qquad \text{one of $x,y\not\in [-L-1,L+1]$} $$
Now suppose $x,y \in [-L-1,L+1]$, then as $|x-y|<\delta \le \delta_0$, we also have
$$ |f(x) - f(y)| \le \epsilon $$
Hence, $f$ is uniformly continuous.
Best Answer
By the mean value theorem, $f(x)-f(y)=f'(\xi)(x-y)$, for some $\xi\in[x,y]$. So, since $f'$ is continuous in $\mathbb{R}$, in a bounded domain $B$ we have $\forall x\in B\ f'(x)\leq C$, for some $C>0$ ($\bar{B}$ must be compact). It follows that $|f(x)-f(y)|\leq C|x-y|$ for all $x,y\in B$.
Next, take $\epsilon >0$. If $|x-y|<\frac{\epsilon}{C}$ then $|f(x)-f(y)|\leq C|x-y|<C\frac{\epsilon}{C}=\epsilon$.