prove the following: using a similar proof as Bolzano's Theorem Let $f : [0, 1] → R$ be continuous with $f (0) = f (1) = 0$. Further suppose that whenever $f(a) = f(b) = 0$ for some $0 \leq a < b \leq 1$ there exists at least one $c \in (a, b)$ such that $f (c) = 0$. Show that $f (x) = 0$ for all $x \in [0, 1]$.
I have this to prove but I'm not sure with what idea to start this proof.
 A: We know that $f(1)=f(0)=0$ and by the property of the function given in the question, then $\exists^{\geq 1} c \in [0,1]$ such that $f(c)=0$.
Define the sets as follow:
$$A:=\{x \in [0,c) | f(x)=0\} \;\; B:=\{x\in (c,1] | f(x)=0\} $$
Since we have $f(0)=0$, then we know $A \neq \emptyset$. Similarly we know $B \neq \emptyset$.
Note that the intervals $[0,c)$ and $(c,1]$ are bounded (above by $c$ and below by $c$ respectively.
Therefore we know that $A$ is bounded above, and $B$ is bounded below. Thus $\exists \sup{A}, \inf{B}$.
Particularly, we have $0 \leq \sup{A} \leq c \leq \inf{B} \leq 1$.
Since $c \in (0,1)$, we have $\sup{A}, \inf{B} \in [0.1]$ and so $f$ has a value for these inputs.
Now, by the approximation property of the supremum, that is: $\exists a \in A \;\text{such that}\;\sup{A}+\frac{1}{n} < a \leq \sup{A}$
Thus, we have:
$$\exists(x_n)_{n \in \mathbb{N}} \subset A \;\;\text{such that}\;\; x_n\rightarrow \sup{A}$$
By continuity, $f(\sup{A}) \rightarrow f(x_n)$. However, by definition of $A$, we have $f(x_n)=0 \;\forall n \in \mathbb{N}$. Thus, $f(\sup{A})=0$.
We do the same procedure for finding $f(\inf{B})$ and then conclude that $f(x)=0 \; \forall x \in [0,1]$ via contradiction of the supremum/infimum properties.
