Exercise 4.5.4

Suppose $F$ is nonempty, let $x,y\in A$ with $x\ne y$ and $f(x)=f(y)$. Pick $z\in (x,y)$ such that $f(z)\ne f(x)$ (if $z$ does not exist $f$ is constant over $(x,y)$ and we are finished early). By the Intermediate Value Theorem every $L\in (f(x),f(z))$ has an $x^{\prime }\in (x,z)$ with $f(x^{\prime })=L$. And since $L\in (f(z),f(y))$ as well we can find $y^{\prime }\in (z,y)$ with $f(y^{\prime })=L$, thus $f(y^{\prime })=f(x^{\prime })$ and so $f$ is not 1-1 at every $L\in (f(x),f(z))$ which is uncountable.