Proving f is a Lipschitz continuous function. (Real analysis)

Suppose that $f: A \to \mathbb{R},$ is such that there is some $M>0$ such that for every $\epsilon>0$, there exists $f_\epsilon: A \to \mathbb{R}$ which is Lipschitz with constant $K(\epsilon)<M$, such that $|f(x)-f_\epsilon(x)|<\epsilon$ for all $x\in A$. Prove that $f$ is Lipschitz continuous.

Also what would be an example of an $f$ which can be approximated by Lipschitz functions like above, except that there is no $M$ and no assumption about the Lipschitz constants $K(\epsilon)$ of the $f_\epsilon$, and for which $f$ is not Lipschitz.

Hint:

\begin{align} |f(x)-f(y)|&=|f(x)-f_\epsilon(x)+f_\epsilon(x)-f_\epsilon(y)+f_\epsilon(y)-f(y)| \\ &\leq |f(x)-f_\epsilon(x)|+|f(y)-f_\epsilon(y)|+|f_\epsilon(x)-f_\epsilon(y)| \end{align}