Subset of $C[0,1]$ is nowhere dense Let $E_n$ be
$$E_n:=\{f\in C[0,1]\mid \text{exist } x_f\in[0,1] \text{ such that } |f(x)-f(x_f)|\leq n|x-x_f|,\, \forall x\in[0,1]\}.$$
How show that $E_n$ is nowhere dense, that is, $\mathrm{int}\  \overline{E_n} = \emptyset$?
Can someone help me?
 A: I haven't checked this thoroughly, but the following proof (informal and in outline form) should work:
First show that $E_n$ is closed: Assume $(f_i)$ is a sequence from $E_n$ that converges to $f$ in $C[0,1]$. For each $i$, choose $x_i\in[0,1]$ verifying  that $f_i\in E_n$. By compactness of $[0,1]$, we may assume $x_n$ converges to some $x\in[0,1]$. Show that $x$ verifies that $f\in E_n$.
Now, to show $E_n$ is nowhere dense, it suffices to show that given $f\in E_n$ and $\epsilon>0$,   there is an $g\notin F_n$ with $\Vert f -g\Vert<\epsilon$. 
To do this, fix $f\in F_n$ and $\epsilon>0$.  First approximate $f$ with a piecewise linear function $\phi$ so that $\Vert \phi-f\Vert<\epsilon/2$. Choose $M$ so that $|\phi'|\le M$, where it exists. Now take a "sawtooth function", $\omega$, with $\Vert \omega\Vert<\epsilon/2$ such that $|\omega'|$ is large. Finally, take $g=\phi+\omega$. This function is piecewise linear and the slope of any "piece" is large in absolute value (as large as you like by choosing the correct sawtooth function). Let's take "large" to mean greater than $n$. It then will follow that given any $x\in[0,1]$, there is an $x_f$ (any point on the same "piece" as $x$ will do) that verifies that $g\notin F_n$. 
And, of course, $\Vert f-g\Vert= \Vert f-(\phi+\omega)\Vert\le\Vert f-\phi\Vert +\Vert\omega\Vert<\epsilon.$
