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Let $I subset R$ be an interval. Let $f : I to R$ be a continuous function.

Assume that $I := [a, b]$. Assume that for all $c, d in [a, b]$ such that $c < d$, there exists $e in [c,d]$ such that $f(e) = f(a)$ or $f(e) = f(b)$. Prove that $f$ is a constant.

Consider this statement: For all $c in [a,b], f(c) in {f(a),f(b)}$

I figured that proving this statement would allow me to prove the function to be constant, but I'm unable to do so. Any thoughts?

We will proceed by contradiction:

I assume you want to prove that the function is constant given the statement

$c in [a, b] Rightarrow f(c) in {f(a), f(b) }$

if that's the case, then we get a continuous function to a discrete space.
If the function takes more than one value, the image of our function will be$ {f(a), f(b)}$ which is disconnected. But continous functions preserve connectedness.

Contradiction Q.E.D.

Let $c$ in $(a,b)$ and let $epsilon$ be a positive real number such that $[c-epsilon,c+epsilon]subseteq[a,b]$. Define a sequence of intervals $(I_n)_{ninmathbb{N}}$ by the formula $I_n=[c-frac{epsilon}{n},c+frac{epsilon}{n}]$. By hypothesis, there exists a sequence of points $(c_n)_{ninmathbb{N}}$ such that $c_nin I_n$ and $f(c_n)in{f(a),f(b)}$. Then $c_n$ tends to $c$ and, by continuity of $f$, we have that $f(c)in{f(a),f(b)}$. Thus, by continuity, $f$ is constant because it takes values in a discrete space.

The set $D:=f^{-1}(f(a))cup f^{-1}(f(b))$ is closed by continuity of $f$ and dense in $I$ by the special property. Clearly $D$ does not intersect the open set $Isetminus D$. By defnition of dense, this means that $Isetminus D$ is empty. Hence $I=D$. This makes $I$ the union of the two non-empty closed sets $f^{-1}(f(a))$, $f^{-1}(f(b))$. As $I$ is connected, these sets must overlap, which means that $f(a)=f(b)$ and ultimately that $f^{-1}(f(a))=I$, i.e., $f$ is constant.

You need to use intermediate value property of continuous functions.

Suppose $f(a) =f(b) $. If $f$ is not constant in $[a, b] $ then there is some value of $f$ which differs from $f(a) =f(b) $ and let's suppose this value $f(c) > f(a) $ (the case $f(c) <f(a) $ is handled in a similar manner). By continuity there is an interval of type $[c-delta, c+delta] $ where all values of $f$ are greater than $f(a) $. And this contradicts the given hypotheses. Thus $f$ must be a constant function.

The case when $f(a) neq f(b) $ leads us to a contradiction as shown next. Let's us assume $f(a) <f(b) $ and choose $k$ such that $f(a) <k<f(b) $. By intermediate value property there is a $cin(a, b) $ for which $f(c) =k$. And by continuity there is an interval of type $[c-delta, c+delta] $ where all values of $f$ lie strictly between $f(a)$ and $f(b) $. This contradiction shows that we must have $f(a) =f(b) $ and as shown earlier the function is constant.