Let $C^\rho(X,x_0)$ be the set of continuous functions $X\to \mathbb R$ with modulus of continuity $\rho$ at $x_0$.
Observation (1): $C^\rho(X,x_0)$ is closed in $C(X)$. If $f$ is in the closure, then:
$$|f(x_0)-f(x)|\leq|f(x_0)-g(x_0)| + |g(x_0)-g(x)| + |g(x)-f(x)|\leq \rho(d(x_0,x)) + 2\sup_{z\in X} |f(z)-g(z)|$$
Where $g\in C^\rho(X,x_0)$ But we can make $\sup_{z\in X} |f(z)-g(z)|$ be arbitrarily small since $f$ is in the closure, so $|f(x_0)-f(x)|\leq \rho(d(x_0,x))$.
Observation (2): If $f,g\in C^\rho(X,x_0)$, then $f-g\in C^{2\rho}(X,x_0)$. This is easy to see.
So, if, for every $\epsilon>0$, we can find an $h\in C(X)$ such that $|h(x)|<\epsilon$ for all $x\in X$ and $h \notin C^{2\rho}(X,x_0)$, then you are done, because for any $f\in C^\rho(X,x_0)$, $f+h\notin C^\rho(X,x_0)$, and therefore $C^\rho(X,x_0)$ is nowhere dense in $C(X)$.
Given $\epsilon>0$, we pick an $x_1\neq x_0$ so that $4\rho(d(x_0,x_1))<\epsilon$. You can find such $x_1$ since $x_0$ is not an isolated point and $\rho(t)\to 0$ as $t\to 0$. Define $\delta=d(x_0,x_1)>0$.
Define $\phi(t)=\frac{\epsilon}{2}(1-\frac{t}{\delta})$ if $t\leq \delta$ and $\phi(t)=0$ if $t>\delta$. Then $h(x)=\phi(d(x_0,x))$ has the property that $|h(x)|<\epsilon$, $h(x_0)=\frac{\epsilon}2$, and $h(x_1)=0$. So $|h(x_0)-h(x_1)|=\frac{\epsilon}2>2\rho(d(x_0,x_1))$. So $h(x)\notin C^{2\rho}(X,x_0)$
Best Answer
It is easy to come up with many examples if you keep in mind that