Show that $U_n$ is a dense subset in $\textsf{C}[0,1]$ I need to show that the open set
$$U_n:= \bigg\{ f\in\textsf{C}[0,1] \,: \sup_{\, 0<|s-t|<1/n} \bigg|\frac{f(s)-f(t)}{s-t}\bigg|>n \bigg\}$$
is dense in $\textsf{C}[0,1]$.
I should use a combination of polynomials and sawtooth waves.
Thanks for your help! 
 A: 
$\textbf{Lemma}$For every continuous $f:[0,1] \to \Bbb{R}$ and $\forall \epsilon>0$ exists a continuous,piecewise linear function $g$ such that $||f-g||_{\infty}<\epsilon$

You can use uniform continuity to prove this lemma.I leave it to you as an exercise.
Now let $f \in C[0,1]$ and $\epsilon>0$.Then exists $g$ as in the lemma such that $||f-g||_{\infty}<\frac{\epsilon}{2}$
It suffices to find $h \in U_n$ such that $||h-g||_{\infty}<\frac{\epsilon}{2}$
$g$ is piecewise linear thus exist $0=t_0<t_1<...<t_N=1$ such that $g$ is linear in $(t_{i-1},t_{i})$
Let $l_i$ the slope of $g$ in $(t_{i-1},t_{i})$.
We define a continuous   function $w:[0,1] \to \Bbb{R}$ such that $0 \leq w(x)< \frac{\epsilon}{2}$ and the slopes of $w$ are greater than $n+\max\{|l_j|:j \in \{1,2,...,N\}\}$
Define $h:=g+w$
Let $x \in [0,1]$. Exists $j \in \{1,2,...,N\}$ such that $x \in [t_{i-1},t_i]$
We chose $|s|<\frac{1}{n}$ small enough such that in the interval with endpoints $x,x+s$,$g,w$ are linear ($s$ can be positive or negative)
If $y \in (x,x+s)$ then $y \in (0,1)$ and $|y-x|<\frac{1}{n}$ and $$|h(y)-h(x)| \geq |w(x)-w(y)|-|g(y)-g(x)|$$ $$\geq (n+\max\{|l_j|:j \in \{1,2,...,N\}\})|y-x|-|l_i||y-x|\geq n|y-x|$$
Thus $h \in U_n$
