Showing a function is continuous on every point in $\mathbb R$ Let $K > 0$ and let $f : \mathbb R  \rightarrow  \mathbb R $satisfy the condition $|f(x) - f(y)|  \leq  K|x - y| $for all $x, y  \in  \mathbb R$. Show that $f$ is continuous at every point $c \in \mathbb R$.
Here's my attempt:

$|f(x) - f(y)|  \leq  K|x - y|  \rightarrow  |f(x) - f(y)|/K \leq  |x
> - y|$
In order for $f$ to have a limit at $c: |x - c| <  \delta   \rightarrow  |f(x) - f(c)| <  \epsilon$
Let $\delta  =  \epsilon$/(2K)$
$|f(x) - f(y)|/K  \leq  |f(x) - f(c)|/K + |f(y) - f(c)|/K <  \epsilon /(2K)
> +  \epsilon /(2K) =  \epsilon /K$
$|f(x) - f(y)| <  \epsilon$

Am I on the right path?
 A: 
Let $\epsilon >0$ be given. Set $\delta= \frac {\epsilon}{K}$ then

Clearly $\delta>0$. Suppose that $x,y\in \mathbb R$ and $|x-y|<\delta$. Then by the given condition on  $f$ we have $|f(x)-f(y)|\le K|x-y|<K\delta=\epsilon$, so $f$ is in fact uniformly continuous on $\mathbb R$>
A: Let $c \in \mathbb{R}$ and let $\varepsilon>0$. We need to find a $\delta>0$ such that $$0<|x-c|<\delta \implies |f(x)-f(c)|<\varepsilon.$$ Choosing $\delta:=\varepsilon/K$ does the trick, since $$|f(x)-f(c)|\leq K|x-c|<K \cdot \frac{\varepsilon}{K}=\varepsilon.$$ Hence $f$ is continuous at $x=c$. Since $c \in \mathbb{R}$ is arbitrary, $f$ is continuous in the real line.
NOTE: A function that satisfies the inequality in your hypothesis is called Lipschitz continuous. 
A: Let $a$ be a real number and let $\varepsilon >0$. By definition of continuity at $x=a$, we need to find $\delta>0$ such that $0<|x-a|<\delta$ implies $|f(x)-f(a)|<\varepsilon$.
Almost done, just put $\delta = \frac{\varepsilon}{K}$, hence $$|f(x)-f(a)| \leq K |x-a| < K \cdot \frac{\varepsilon}{K}=\varepsilon.$$
Since $a \in \mathbb R$ is arbitrary, the $f$ is continuous on $\mathbb{R}$.
