Show that $f$ is uniformly continuous given that $f'$ is bounded. Let $D\subset \mathbb{R}$ be an interval and $f:D\rightarrow \mathbb{R}$ a continuously differentiable function such that its derivative $f':D\rightarrow \mathbb{R}$ is bounded. Show that $f$ is uniformly continuous.
I will admit before I begin; I am stabbing in the dark.
If we restrict the domain of $f$ to some interval $[-X,X] \subset D$, then $f:[-X,X]\rightarrow \mathbb{R}$ is uniformly continuous. 
Now since we know that $f'$ is bounded we can say that for some arbitrary $y \in D$, $|f'(x)| = \lim_{x \rightarrow y}\frac{|f(x) - f(y)|}{|x-y|} \leq K   ( \forall x \in D:x \neq y) $ for some $K >0 $
This is not going to sounds mathematical, but I need to somehow get rid of the limit so I can proceed with something like $\frac{|f(x) - f(y)|}{|x-y|} \leq K$

Let $f:[-X,X]\rightarrow \mathbb{R}$ has the conditions that you want. Let $\epsilon>0$, then we want to find $\delta>0$ such that:
$$\forall x,y\in [-X,X](|x-y|<\delta \rightarrow |f(x)-f(y)|<\epsilon)$$
By Mean value theorem for every $x,y\in [-X,X],x<y$, there exists $c\in (-X,X)$ such that
$$\frac{f(y)-f(x)}{y-x}=f'(c)$$ 
Because for every $x\in (-X,X)$, we have $|f'(x)|<K$, therefore for every $x,y\in [-X,X],x<y$
$$\frac{|f(y)-f(x)|}{|y-x|}<K$$
so we choose $\delta = \frac{\epsilon}{K}$
