Is the set of continuous maps on $[0,1]$ with finitely many roots open in the $C^0$ topology?

Question

Let I would like to show that the set $$S=\{f\in\mathcal{C}([0,1],\mathbb{R}):f\text{ has finitely many zeros}\}$$ is open. By considering $f_n(x)=\frac{x}{n}$, we see that $S$ is not closed. By applying the transversality lemma to $F(x,\theta)=f(x)+\theta$, we find that $S$ is dense. I am hoping to show that $S$ is open. I know that if $f$ is transverse to $\{0\}$ and $g\in\mathcal{C}([0,1]^2,\mathbb{R})$, then there exists an $\varepsilon_g$ such that $f+\varepsilon g$ is tranverse to $\{0\}$ for every $\varepsilon\le\varepsilon_g$. However, this is not good enough to show openness of $S$ because $\varepsilon_g$ depends on the direction $g$. I found an infinite-dimensional version of the transversality theorem here that would show that $S$ is open but I am hoping there is an approach with a simpler set of conditions to check than the one in the link.

Answer 1

No, the set $S$ is not open!

Consider say the function $f(x)=x$ for all $x\in I:=[0,1]$, and any arbitrary $\varepsilon\in(0,\frac12)$. Note that the function $g\in C^0(I)$ defined by $$ g(x)=\begin{cases} 0 & x\in [0,\varepsilon]\\ 2(x-\varepsilon) & x\in [\varepsilon,2\varepsilon]\\ x & x\in[2\varepsilon,1] \end{cases} $$ satisfies $\lVert f-g\rVert_{C^0(I)}=\varepsilon$ and $g\notin S$.