Compactness and continuity problems 
Let $f: K \to \mathbb{R}$ be a continuous function from compact set $K$ (so $f$ is uniformly continuous). Then show for $\forall \epsilon >0$, $\exists A >0$ (depending on $\epsilon)$ such that $$|f(x) - f(y)| \leq A|x-y| + \epsilon,$$ for $x,y \in K$(Hint: argue by contradiction)

Okay so I have some ideas, but I didn't feel like they were going anywhere. 
To prove statement, assume that $\forall \epsilon > 0$, $\nexists A >0$ such that the inequality holds or $\forall \epsilon > 0$, $\forall A >0$ is inequality is false. Then arrive at some contradiction.
So I decided to take $A = \epsilon$. But that wasn't going anywhere. Can I just get a hint? No answer please. Note that I think it is better to consider the case $|f(x)| \leq A|x| + \epsilon$ by setting $f(0) = 0$ as I think the general case in here can be replaced by a translation.
 A: Assume for contradiction, that there exists $\varepsilon_0 > 0$ such that for any $n\in \mathbb{N}$ there exists $x_n, y_n \in K$ satisfying
$$
\tag{1} |f(x_n) - f(y_n) | \geq n |x_n - y_n| + \varepsilon_0.
$$
Using the compactness of $K$ we extract a converging subsequence from $\{x_n\}$ and $\{y_n\}$. Namely, let $n_k$ be a subsequence of $\mathbb{N}$ such that $x_{n_k} \to x_0$ and $y_{n_k} \to y_0$. Since $f$ is continuous on $K$ and $K$ is compact, it follows that $f$ is bounded. Fix $M>0$ such that $|f(x)| \leq M$ for all $x\in K$.  From $(1)$ we get
$$
2M \geq n_k |x_{n_k} - y_{n_k}| + \varepsilon_0.
$$
Since $n_k |x_{n_k} - y_{n_k}|$ is bounded above, and $n_k \to \infty$, it follows that $|x_{n_k} - y_{n_k}| \to 0$. But then, the continuity of $f$ implies $f(x_{n_k}) - f(y_{n_k}) \to 0$ and passing to the limit in $(1)$ along $\{n_k\}$ we get
$$
0 \geq \limsup\limits_{k\to \infty} n_k |x_{n_k} - y_{n_k}| + \varepsilon_0 \geq \varepsilon_0 > 0
$$ 
which is a contradiction.
A: The rough Lipschitz condition does not give lot of control when $x$ and $y$ are close. May start by noticing that for $|x-y|>\delta$, there exists a bounded function $\omega:[\delta,diam(K)]\to \mathbb{R}$ such that $|f(x)-f(y)|<\omega(|x-y|)$. 
