Show that the subset of strictly monotonous functions in $C(I)$ is nowhere dense 
Show that the subset of strictly monotonous functions in $C(I)$ is nowhere dense in $C(I)$

My idea is proving that the closure of the subset is the subset of weakly monotonous functions, and then it will be easy to prove that the interior of this subset is empty. It's the first part that I'm having trouble with. 
I can't seem to easily show that a weakly monotonous function has a strictly monotonous in each environment. I have a vague idea for showing that when the number of constant intervals is finite, but not in the general case.
Also showing that every element of the closure is weakly monotonous seems hard. I don't know how to approach it.
 A: This is based upon a common alternative characterization of convexity which is used to show that the pointwise limit of strictly convex functions is, at least, convex.
For a fixed function $f$ on an interval $I$, let's define
$$f^{[1]}(x,y)=\frac{f(x)-f(y)}{x-y}$$
for $x$ and $y$ distinct. Then you may verify that $f$ is strictly increasing if and only if
$$f^{[1]}(x,y)>0$$
for every $x$ and $y$ distinct. Now suppose that $\{f_n\}$ is strictly increasing for every $n$ and that this sequence converges pointwise to $f$ on $I$. Pick $x$ and $y$ distinct from $I$. Then
$$f^{[1]}(x,y)=\frac{f(x)-f(y)}{x-y}=\lim_{n\rightarrow\infty}\frac{f_n(x)-f_n(y)}{x-y}\geq 0\,.$$
which demonstrates that $f$ is (weakly) increasing. As a last remark, if $\{f_n\}$ converges in $C(I)$ (with the sup norm, I assume) then of course it converges pointwise to its limit function.
A: If a sequence $f_n$ of weakly monotone functions on $I$  converges to $f$ pointwise on $I,$ then $f$ is weakly monotone. Proof: Let $A= \{n: f_n \text { is weakly increasing}\}, B= \{n: f_n \text { is weakly decreasing}\}.$ One of $A,B$ is infinite. Suppose WLOG it's $A.$ Then $f$ is the pointwise limit of weakly increasing functions on $I.$ Hence $f$ is weakly increasing. (The last claim follows from a simple result for number sequences: If $a_n, b_n$ are convergent real sequences and $a_n \le b_n$ for all $n,$ then $\lim a_n \le \lim b_n.$)
Let $X$ be the set of weakly monotone functions in $C(I).$ The above shows that $X$ is closed in $C(I).$ As you seem to know, $X$ is nowhere dense in $C(I).$ This proves your result, since the closure in $C(I)$ of the set of strictly monotone functions in $C(I)$ is contained in $X.$
