I am struggling with the following uniform-continuity proof:
Let $c \in (0,1)$. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function with the property that for all $x,y \in \mathbb{R}$, we have that $|f(x)-f(y)| = c|x-y|$. Then $f$ is uniformly continuous.
I know that to show a function is uniformly continuous, you need to apply the $\epsilon$ and $\delta$-ball definition to any two arbitrary points and their outputs. I just don't know how a proof like this would begin, and then finish. What am I trying to show here?
Well, first let us see if a function $f$ is uniformly continuous, what does it mean?
$$\forall \epsilon > 0, \exists \delta > 0 \text{ such that } \forall x, y \in \mathbb{R} \text{ with } \left| x - y \right| < \delta, \text{ we get } \left| f \left( x \right) - f \left( y \right) \right| < \epsilon$$
Now, let us see what would happen if our function $f$ is uniformly continuous. Then, we would get
$$\left| f \left( x \right) - f \left( y \right) \right| = c \left| x - y \right| < \epsilon$$
for any $x, y \in \mathbb{R}$. Therefore, we can also have (since $c > 0$) $\left| x - y \right| < \dfrac{\epsilon}{c}$.
Now, we write the formal ("textbook") proof for the uniform continuity.
Let $\epsilon > 0$ be given. Choose $\delta = \dfrac{\epsilon}{c}$ (where $c$ comes from the property of $f$ given to us). Therefore, $\forall x, y \in \mathbb{R}$ with the property that $\left| x - y \right| < \delta = \dfrac{\epsilon}{c}$, we have
$$\left| f \left( x \right) - f \left( y \right) \right| = c \left| x - y \right| < c \dfrac{\epsilon}{c} = \epsilon$$
Hence, $f$ is uniformly continuous.