Proposition: a function $f: \mathbb{R} \to \mathbb{R}$ is uniformly continuous if and only if it is Lipschitz continuous.
Proof:
Suppose $f$ is uniformly continuous. Then, by the definition of uniform continuity for every $\epsilon > 0$ there is $\delta > 0$ such that $|x-y|<\delta$ implies that $|f(x)-f(y)|<\epsilon$. Given an $\epsilon >0$, we can write $\delta(\epsilon)$ to indicate that $\delta$ depends on $\epsilon$. Let $$ K = max_{\epsilon>0}\frac{\epsilon}{\delta(\epsilon)}$$ Then for every $x, y, |f(x)-f(y)|<K|x-y|$. Thus by the definition of K we have that for every $\epsilon >0, K\geq \frac{\epsilon}{\delta(\epsilon)}$ and so $\epsilon \leq K\delta(\epsilon)$, and so when $|x-y| = \delta(\epsilon)$ then $$|f(x)-f(y)| ]\leq \epsilon \leq K\delta(\epsilon) = K|x-y|$$ and so f is Lipschitz with constant K.
On the other hand, suppose that $f$ is Lipschitz with constant $K$, i.e. $|f(x)-f(y)|\leq K|x-y|$. Pick any $\epsilon >0$ and let $\delta = \epsilon/K$. Then if $|x-y|<\delta$, we have that $$|f(x)-f(y)| \leq K|x-y| \leq K\delta = \epsilon, $$ and so f is uniformly continuous.
Should the proposition instead be that if $f$ is lipschitz then it is uniformly continuous?