Suppose $f:\mathbb{R} \to \mathbb{R}$ is uniformly continuous. Show that $f(x+1)-f(x)$ is bounded Suppose $f:\mathbb{R} \to \mathbb{R}$ is uniformly continuous. Show that $f(x+1)-f(x)$ is bounded.
I have considered the question, and my current approach is to show that there exists $\delta > 1$ such that there is some fixed $\epsilon > 0$ for which $|x-p| < \delta \implies |f(x)- f(p)| < \epsilon$, thus bounding $f(x+1)-f(x)$ by this $\epsilon$. Is my current approach correct, and are there any hints or alternative approaches to this question?
Thanks in advance.

By definition, there exists $N\in\mathbb{N}$ such that $y\in (x-N^{-1},x+N^{-1})$ implies
$$|f(y)-f(x)|<1$$
This same logic applies to $x_k=x+\frac{k}{2N}$:
$$|f(y)-f(x_k)|<1$$
Then
$$ |f(x+1)-f(x)|=\left|f\left(x+\frac{2N}{2N}\right)-f(x)\right|$$
$$=\left|\sum_{k=1}^{2N}\left[f\left(x+\frac{k}{2N}\right)-f\left(x+\frac{k-1}{2N}\right)\right]\right|$$
$$\leq \sum_{k=1}^{2N}\left|f\left(x+\frac{k}{2N}\right)-f\left(x+\frac{k-1}{2N}\right)\right|$$
Since
$$\frac{k}{2N}-\frac{k-1}{2N}=\frac{1}{2N}<\frac{1}{N}$$
we may conclude that the whole sum is bounded by $|f(x+1)-f(x)|<2N$.
