Modulus of Continuity, Take 2 This is a follow-up to my last question: Modulus of Continuity.  I accidentally asked the wrong question there, so I'm going to start over and hopefully ask the right question.
I'll repeat the relevant definitions.  Let $\rho: \mathbb{R}^+ \to \mathbb{R}^+$ be a continuous nondecreasing function such that $\rho(t) = 0$ if and only of $t = 0$.  If you can answer my question in the special case $\rho(t) = Ct$ where $C$ is a constant then it will probably be possible to adapt your construction to the general case.
Say that a function $f: X \to \mathbb{R}$ on a metric space has modulus of continuity $\rho$ at a point $x_0 \in X$ if $|f(x) - f(x_0)| \leq \rho(d(x,x_0))$ for every $x \in X$.  For example, a function has modulus of continuity $Ct$ at $x_0$ if and only if it is Lipschitz with Lipschitz constant $C$ at $x_0$.  
Question If $X$ is a compact metric space without isolated points, is it true that the set of all continuous functions on $X$ which have modulus of continuity $\rho$ at some point of $X$ is nowhere dense in $C(X)$ equipped with the supremum norm?
To prove that the answer is affirmative for a given $X$ one must be able to construct functions of arbitrarily small norm which oscillate arbitrarily rapidly.  For example, if $\rho(t) = Ct$ and $X = [0,1]$ then one can use a piecewise linear function such that the slope of each linear piece is larger than $C$ in absolute value.  However, I don't see how to generalize this idea to an arbitrary compact metric space without isolated points.
 A: A detailed answer was posted by fedja on MathOverflow. His main points were: 


*

*the set of functions that are $\rho$-continuous somewhere is not nowhere dense. In fact it is dense, because every function can be flattened a bit: namely,  $(f-a-\epsilon)^+-(f-a)^-+a$ is within $\epsilon$ of $f$ but is flat where $f$ took values between $a$ and $a+\epsilon$.

*the set of functions that are $\rho$-continuous somewhere is of first category. The key steps are summarized below.

Fix  a constant $A$ and consider the set $F_A$ of functions $f$ such that there exists a point $x\in X$ at which $f$ has global modulus of continuity $A\rho $. This set is closed: if $f_n\in F_A$ converge to $f$ uniformly and $x_n\in X$ are the corresponding bad points, then any accumulation point $x$ of the sequence $x_n$ is bad for $f$. Also, it contains no open set, because  any function $f \in C(X)$ can be perturbed by small spikes placed along some $\delta$-net in $X$ so as to lose the membership in $F_A$.
A: Try functions such as
$f(x) = \min(\epsilon, 2 \rho(d(x,x_0)))$.  
