Nowhere dense set.... Let $A_n$ be a subset of continuous functions on $[0,1]$ given by:
$A_n$ = {$f∈C[0,1]$:there exists $x∈[0,1]$ such that $|f(x)−f(y)|≤n|x−y|$ for all $y∈[0,1]$}.
Show $A_n$ is nowhere dense, and use this fact to show that there are nowhere differentiable continuous functions on $[0,1]$.
We already showed $A_n$ is closed. How to show interior of $A_n$ is empty?
 A: As it's currently written, this answer provides only the idea of the proof.
A function $f(x)=0$ belongs to $A_n$, so $A_n\ne\emptyset$. Now let's take a function $$g(x)=1-2|x-1/2|,$$
a "hat" function. Let's define it by periodicity to the whole $\Bbb R$: $$g(x)=g(x-1)=g(x+1)\quad \forall x\in \Bbb R.$$
Now take $$g_k = \frac {1}{2^k} g(2^kx)$$and$$G_N(x)=\sum_{k\ge N}g_k(x).$$
For large $N$ this function is close to zero (and, therefore, to $A_n$), it is continuous as an absolutely convergent series of continuous functions, but it can't belong to $A_n$. 
We want to show that $$\forall f\in A_n\quad f+G_N\notin A_n.$$
(to be continued)
A: I'll write something to show that there are nowhere differentiable functions on $C[0,1]$.
Proof:
\begin{align*}
A_n\text{ nowhere dense }&\iff\left(\overline{A_n}\right)^\circ=\varnothing\\
&\iff X\setminus(\overline{A_n})^\circ=X\\
&\iff\overline{X\setminus \overline{A_n}}=X\\
&\iff\overline{(X\setminus A_n)^\circ}=X\\
&\iff (X\setminus A_n)^\circ\text{ is dense in }X\\
&\iff(X\setminus A_n)\text{ contains a dense open subset}.
\end{align*}
Let $G_n=(X\setminus A_n)^\circ$, then ${G_n}\subset(A_n)^c$
By Baire's category theorem, $\cap_nG_n$ is dense in $X$ and hence $\cap_nG_n\neq\emptyset$.
For any $f\in{\cap_nG_n}$, since $f\in(A_n)^c$ for any n,
$|f(x)-f(y)|\ge n|x-y|$ for all $x,y\in[0,1]$ and all n.
Thus $f$ is nowhere differentiable.
