Using Bounded Derivative to Show Uniform Continuity

Question

Suppose $f: (a,b) \to \mathbb{R}$ is differentiable on $(a,b)$ and $|f '(x)|\leq M$ for all $x \in (a,b)$. Prove that $f$ is uniformly continuous on $(a,b)$. My attempt: I started by observing the definition of uniform continuity, which is for all $\epsilon \gt 0$ there exists $\delta \gt 0$ such that for all $x,y \in (a,b)$, $|x-y|\lt \delta$ implies $|f(x) - f(y)|\lt \epsilon$. Using this I began by trying to find $\delta$. My first thought was to use the information about the derivative. I wrote out that $\lim_{x\to y}$$|$$(f(x) - f(y) \over x - y)$|$\leq M$. Using this, $-M$($\lim_{x\to y}(x - y))$$\leq$$\lim_{x \to y}(f(x) - f(y))$$\leq$$M(\lim_{x \to y}(x - y))$. Now this implies that $|\lim_{x\to y}f(x) - f(y)|$$\leq$$M(\lim_{x \to y}(x - y))$. I'm thinking that this is what will get me to the $\delta$ that I'm looking for. But the limit on the right hand side approaches 0, so this should imply that $|\lim_{x \to y}f(x) - f(y)|$$\leq$ $0$. Is this the result that will show uniform continuity?