Lipschitz Continuous $\Rightarrow$ Uniformly Continuous The Question: Prove that if a function $f$ defined on $S \subseteq \mathbb R$ is Lipschitz continuous then $f$ is uniformly continuous on $S$. 

Definition. A function $f$ defined on a set $S \subseteq \mathbb R$ is said to be Lipschitz continuous on $S$ if there exists an $M$ so that $$\frac{|f(x) - f(c)|}{|x - c|} \le M$$ for all $x$ and $c$ in $S$ such that $x \ne c$. 

My Heuristic Interpretation: if $f$ is Lipschitz continuous then the "absolute slope" of $f$ is never unbounded i.e. no asymptotes. 

Definition. A continuous function $f$ defined on $\mathrm{Dom}\, (f)$ is said to be uniformly continuous if for each $\varepsilon > 0 \ \exists \ \delta > 0$ s.t. $\forall \ x, c \in \mathrm{Dom}\, (f)$ $$ |x - c| \le \delta \ \Rightarrow \ |f(x) - f(c)| \le \varepsilon$$


Proof:
$f$ Lipschitz continuous $\Rightarrow$ $|f(x) - f(c)| \le M|x - c|$. Since we suppose $|x - c| \le \delta$ for uniform continuity, we have $x$ within $\delta$ of $c$, so $|x| \le |c| + \delta$. So taking $\delta = \varepsilon/M$
\begin{align*}
|f(x) - f(c)| &\le M|x - c| \\
& \le M\delta \\
& = \varepsilon
\end{align*}
My Question: Is my proof valid with the assumptions taken? 
 A: It’s not very well organized, and it has some extraneous clutter, but it also has the core of the argument. You want to show that for each $\epsilon>0$ there is a $\delta>0$ such that $|f(x)-f(c)|<\epsilon$ whenever $x,c\in\operatorname{dom}f$ and $|x-c|<\delta$, so in a polished version of the argument your first step should be:

Suppose that $f$ is Lipschitz continuous on some set $S$ with Lipschitz constant $M$, and fix $\epsilon>0$.

You’ve already worked out that $\epsilon/M$ will work for $\delta$, so you can even start out with:

Fix $\epsilon>0$ and let $\delta=\frac\epsilon{M}$.

Now you want to show that this choice of $\delta$ does the job.

Clearly $\delta>0$. Suppose that $x,c\in S$ and $|x-c|<\delta$. Then by the Lipschitz continuity of $f$ we have $|f(x)-f(c)|\le M|x-c|<M\delta=\epsilon$, so $f$ is uniformly continuous on $S$. $\dashv$

Added: Your heuristic interpretation of Lipschitz continuity is inaccurate enough that it may well lead you astray at some point. Consider the function
$$f(x)=\begin{cases}
x\sin\frac1x,&\text{if }x\ne0\\
0,&\text{if }x=0\;.
\end{cases}$$
This function has no vertical asymptotes, but it’s not Lipschitz continuous: for $n\in\Bbb Z^+\setminus\{0\}$ we have
$$\begin{align*}
\left|\frac{f\left(\frac1{2n\pi}\right)-f\left(\frac1{2n\pi+\frac{\pi}2}\right)}{\frac1{2n\pi}-\frac1{2n\pi+\frac{\pi}2}}\right|&=\left|\frac{\frac1{2n\pi+\frac{\pi}2}}{\frac1{2n\pi}-\frac1{2n\pi+\frac{\pi}2}}\right|\\
&=\left|\frac1{\frac{2n\pi+\frac{\pi}2}{2n\pi}-1}\right|\\
&=\left|\frac{2n\pi}{\pi/2}\right|\\
&=4|n|\,,
\end{align*}$$
which can be made as large as you want. This function has very, very steep bits, but they’re also very, very short.
A: I suppose one small detail to add to Brian's answer is that in the argument we assumed $M\neq 0$. But the case for $M=0$ is trivial enough as that means $f$ is constant, which is clearly uniformly continuous (setting $\delta=\epsilon$).
$\textbf{Edit:}$
Suppose $f:S\to\mathbb{R}$ is Lipchitz continuous, then we have some $M\geq 0$ such that for all $x,y\in S$,
$$|f(x)-f(y)|\leq M|x-y|$$
Suppose $M>0$. Then, given some $\epsilon>0$, we set $\delta=\frac{\epsilon}{M}$. Then for all $x,y\in S$, we have
$$|x-y|<\delta\implies |f(x)-f(y)|\leq M|x-y|<M\cdot \frac{\epsilon}{M}=\epsilon$$
Hence $f$ is uniformly continuous.
If $M=0$, then it is easy to see $f(x)=f(y)$ for all $x,y\in S$, i.e. $f$ is constant. One can check easily that this is uniformly continuous (setting $\delta=\epsilon$).
Note, however, that the converse is not true. For instance, one can check that the function $f:[0,\infty)\to\mathbb{R}$ given by $f(x)=\sqrt{x}$ is uniformly continuous but not Lipschitz continuous.
