Let $A_{n}:=\{f \in C([0,1]): \exists x \in [0,1], \forall y \in [0,1] \operatorname{so that} |f(x)-f(y)|\leq n |x-y|\}$
I have already shown that $A_{n}$ is closed w.r.t $(C([0,1]), \vert \vert \cdot \vert \vert_{\infty})$ but now I want to show that $A_{n}$ is nowhere dense, and I am given the hint of considering the linear function $s:[0,1] \to [0,C]$ with slope $\pm(n + D)$ where $D$ and $C$ are some constants $> 0$.
$(\overline{A_{n}})^{c}=(A_{n})^{c}:=\{f \in C([0,1]): \forall x \in [0,1], \exists y\in[0,1] \operatorname{so that} |f(x)-f(y)|> n|x-y|\}$
Let $g \in C([0,1])$ and $ \epsilon > 0$: since $g$ is continuous for all $x \in [0,1]$ then there exists $\delta_{x} > 0$ so that for all $|x-y|< \delta_{x}\Rightarrow |g(x)-g(y)|<\epsilon$
$\vert \vert g - f \vert \vert_{\infty}=\sup_{x \in [0,1]}|g(x)-f(x)|$
I am unsure on how to use the linear function $s$. Any ideas?