This is the exercise 4.b of chapter 8 from Spivak:
Suppose that $f$ is continuous on $[a,b]$ and that $f(a)<f(b)$. Prove that there are numbers $c$ and $d$ with $a \leq c <d \leq b$ such that $f(c)=f(a)$ and $f(d)=f(b)$ and $f(c)<f(x)<f(d)$ for all $x$ in $(c,d)$.
So here's my attempt at a proof:
First consider the set $A=\{x \in [a,b] : f(x)=f(a)\}$. Since $A \neq \emptyset$ ($a$ is in $A$) and $b$ is an upper bound for this set, it follows that $A$ has a least upper bound $c=\sup A$. Let's show that $f(c)=f(a)$. Suppose that $f(c) \neq f(a)$, then there are two possibilities, $f(c)>f(a)$ or $f(c)<f(a)$. If $f(c)>f(a)$, by continuity of $f$, there exists a number $\delta>0$ such that if $c-\delta<x<c+\delta$, then $f(x)>f(a)$, but by definition of least upper bound, there is $y \in A$ such that $c-\delta<y<c+\delta$, so $f(y)>f(a)$, which is absurd since $y$ is an element of $A$. Now suppose $f(c)<f(a)$, by the same argument used in the previous case we arrive to a contradiction. In conclusion, we must have $f(c)=f(a)$.
Now define $B=\{x \in [a,b]:f(x)=f(b)\}$. Using analogous arguments to the two statements related $A$, it follows that there exists a greatest lower bound $d=\inf B$ and that $f(d)=f(b)$.
So it remains to prove the last part of the exercise: for all $y$ such that $c<y<d$, we have $f(c)<f(y)<f(d)$. I couldn't prove this statement, I would really appreciate some help. Thanks in advance.
If it were not the case that for all x between $c$ and $d$, $f(c) < f(x) < f(d)$, then there would be a point $x$ between $c$ and $d$ with $f(x) \leqslant f(c) = f(a)$ or $f(x) \geqslant f(d) = f(b)$.
By continuity and the intermediate value theorem, there would be a point $y$ between $c$ and $d$, where $f(y) = f(a)$ or $f(y) = f(b)$, contradicting your sup/inf construction.