Let $$f\in \mathcal C^0(\mathbb R),\quad f(0)=0,\quad f(a)>0,\quad f(-a)<0,\quad a>0.$$
Let interval $I$ satisfies: $$\{x\in I\ |\ f(x)=0\}=K,$$ where $K$ is a subinterval, $K\subset I,$ and $0\in K$.
Question: Does such an $I$ always exist for any $f$?
Yes, of course. Let $A=\{x>0|f(x)\neq 0\}$. $A$ must be nonempty. Take its element, $a^*$, to be the closest element to $0$.
Similarly, let $A'=\{x<0|f(x)\neq0\}$. $A'$ must be nonempty. Take its element, $a'^*$, to be the closest element to $0$. $(a'^*,a^*)$ is the interval we want.
No. Intuitively, it is easy to construct Weierstrass function $f$ such that $\forall \varepsilon>0$, we have $\{x\in (0,\varepsilon)|f(x)=0\}$ is infinite and discrete.
I think $(2)$ is correct (so $(1)$ is wrong), because we can construct a $f$ such that $A$ is a discrete set without minimum, and $\inf (A)=0$. Does this make sense?
A related question: Does there exist an interval with singleton interior (here "interior" means topological interior)?