Intro to Proofs: Continuity Let $c > 0$ and $f : \mathbb{R} → \mathbb{R}$ satisfy
$$|f(x) − f(y)| ≤ c|x − y|$$
for all $x, y ∈ \mathbb{R}$.Show that $f$ is continuous.
Does showing that $|x − y|≤ \frac{\delta}{c}$ and then setting $|f(x) − f(y)|/c ≤ \frac{\delta}{c}$ show that f is continuous?  Therefore since we have a delta for any epsilon, that f must be continuous? 
 A: You should state things more clearly. 
You get the idea but your reasoning can be more clear.
Let $\epsilon > 0$ (always start this way!).
You look for:
$$
 |f(x) − f(y)| \le\epsilon
$$
for each $y$ such as
$$
|y-x|\le \delta(\epsilon,x)
$$
As you said:
$$ |f(x) − f(y)|\le c|x-y|
$$so the first equation is true when
$$
 c|x-y|\le \epsilon
\iff |x-y|\le \frac\epsilon c
$$
hence taking $\delta(\epsilon,x) = \frac\epsilon c$ is fine and $f$ is continuous.
A: You are on the right track.
For every $\epsilon > 0$, take $\delta = \frac{\epsilon}{c}$. Then we have 
$$\begin{align}&|x - y| < \delta \Rightarrow |f(x) - f(y)| \leq c|x - y| < c\  \delta = c\ \Big( \frac{\epsilon}{c}\Big) = \epsilon\\\Rightarrow &|x - y| < \delta \Rightarrow |f(x) - f(y)| < \epsilon\end{align}$$
For every $x,y \in \mathbb{R}$. Then $f$ is continuous. 
And even further, as long as it doesn't depend on $x$ and $y$ taken in $\mathbb{R}$, the function is uniformly continuous.
What is between the lines here is: Every Lipschitz function is uniformly continuous.  
