Exercise 4.5.1

Since $[a,b]$ is connected, so too must be $f([a,b])$, which must be an interval containing both $f(a)$ and $f(b)$; therefore $[f(a),f(b)]$ (or $[f(b),f(a)]$ if $f(b)<f(a)$) is a subset of $f([a,b])$, completing the proof.