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:
$$\frac{f\left(\frac1{2n\pi}\right)-f\left(\frac1{2n\pi+\frac{\pi}2}\right)}{\frac1{2n\pi}-\frac1{2n\pi+\frac{\pi}2}}=\frac{\frac1{2n\pi+\frac{\pi}2}}{\frac1{2n\pi}-\frac1{2n\pi+\frac{\pi}2}}=\frac1{\frac{2n\pi+\frac{\pi}2}{2n\pi}-1}=\frac{2n\pi}{\pi/2}=4n\;,$$
which can be made as large as you want. This function has very, very steep bits, but they’re also very, very short.
Best Answer
Following @Sujit's
nearlycorrect outline, suppose $f$ is Lipschitz continuous: $$|f(x) - f(y)| < L|x-y|$$and let $\epsilon > 0$ be arbitrary and choose $\color{red}{\delta = \frac{\epsilon}{L}}$. Then $$\begin{align} |x-y| < \delta &\implies |x-y| < \frac{\epsilon}{L}\\ & \implies L|x-y| < \epsilon\\ & \implies \underbrace{|f(x) - f(y)| < L|x-y|}_{supposition} < \epsilon\\ & \implies |f(x) - f(y)| < \epsilon \end{align}$$ $$ \square$$